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feat: add solutions to lc problem: No.0114
No.0114.Flatten Binary Tree to Linked List
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‎solution/0100-0199/0114.Flatten Binary Tree to Linked List/README.md

+101-37
Original file line numberDiff line numberDiff line change
@@ -53,6 +53,14 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：寻找前驱节点**
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先序遍历的访问顺序是“根、左子树、右子树”，左子树最后一个节点访问完后，接着会访问根节点的右子树节点。
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因此，对于当前节点，如果其左子节点不为空，我们找到左子树的最右节点，作为前驱节点，然后将当前节点的右子节点赋给前驱节点的右子节点。然后将当前节点的左子节点赋给当前节点的右子节点，并将当前节点的左子节点置为空。然后将当前节点的右子节点作为下一个节点，继续处理，直至所有节点处理完毕。
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时间复杂度 $O(n)$，空间复杂度 O(1)$。其中 $n$ 是树中节点的个数。
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<!-- tabs:start -->
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### **Python3**
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# self.left = left
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# self.right = right
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class Solution:
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def flatten(self, root: TreeNode) -> None:
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def flatten(self, root: Optional[TreeNode]) -> None:
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"""
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Do not return anything, modify root in-place instead.
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"""
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}
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```
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### **TypeScript**
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```ts
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/**
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* Definition for a binary tree node.
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* class TreeNode {
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* val: number
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* left: TreeNode | null
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* right: TreeNode | null
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* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
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* this.val = (val===undefined ? 0 : val)
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* this.left = (left===undefined ? null : left)
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* this.right = (right===undefined ? null : right)
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* }
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* }
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*/
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/**
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Do not return anything, modify root in-place instead.
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*/
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function flatten(root: TreeNode | null): void {
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while (root != null) {
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if (root.left != null) {
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let pre = root.left;
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while (pre.right != null) {
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pre = pre.right;
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}
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pre.right = root.right;
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root.right = root.left;
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root.left = null;
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}
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root = root.right;
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}
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}
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```
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### **C++**
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```cpp
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### **Go**
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```go
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/**
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* Definition for a binary tree node.
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* type TreeNode struct {
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* Val int
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* Left *TreeNode
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* Right *TreeNode
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* }
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*/
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func flatten(root *TreeNode) {
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for root != nil {
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if root.Left != nil {
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pre := root.Left
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for pre.Right != nil {
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pre = pre.Right
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}
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pre.Right = root.Right
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root.Right = root.Left
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root.Left = nil
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}
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root = root.Right
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}
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}
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```
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```go
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/**
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* Definition for a binary tree node.
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}
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```
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### **TypeScript**
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```ts
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/**
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* Definition for a binary tree node.
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* class TreeNode {
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* val: number
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* left: TreeNode | null
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* right: TreeNode | null
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* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
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* this.val = (val===undefined ? 0 : val)
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* this.left = (left===undefined ? null : left)
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* this.right = (right===undefined ? null : right)
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* }
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* }
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*/
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/**
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Do not return anything, modify root in-place instead.
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*/
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function flatten(root: TreeNode | null): void {
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while (root != null) {
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if (root.left != null) {
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let pre = root.left;
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while (pre.right != null) {
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pre = pre.right;
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}
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pre.right = root.right;
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root.right = root.left;
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root.left = null;
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}
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root = root.right;
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}
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}
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```
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### **JavaScript**
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```js
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/**
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* Definition for a binary tree node.
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* function TreeNode(val, left, right) {
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* this.val = (val===undefined ? 0 : val)
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* this.left = (left===undefined ? null : left)
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* this.right = (right===undefined ? null : right)
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* }
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*/
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/**
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* @param {TreeNode} root
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* @return {void} Do not return anything, modify root in-place instead.
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*/
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var flatten = function (root) {
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while (root) {
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if (root.left) {
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let pre = root.left;
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while (pre.right) {
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pre = pre.right;
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}
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pre.right = root.right;
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root.right = root.left;
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root.left = null;
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}
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root = root.right;
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}
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};
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```
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### **...**
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```

‎solution/0100-0199/0114.Flatten Binary Tree to Linked List/README_EN.md

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@@ -58,7 +58,7 @@
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# self.left = left
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# self.right = right
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class Solution:
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def flatten(self, root: TreeNode) -> None:
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def flatten(self, root: Optional[TreeNode]) -> None:
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"""
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Do not return anything, modify root in-place instead.
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"""
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}
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```
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### **TypeScript**
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```ts
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/**
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* Definition for a binary tree node.
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* class TreeNode {
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* val: number
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* left: TreeNode | null
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* right: TreeNode | null
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* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
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* this.val = (val===undefined ? 0 : val)
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* this.left = (left===undefined ? null : left)
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* this.right = (right===undefined ? null : right)
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* }
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* }
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*/
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/**
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Do not return anything, modify root in-place instead.
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*/
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function flatten(root: TreeNode | null): void {
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while (root != null) {
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if (root.left != null) {
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let pre = root.left;
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while (pre.right != null) {
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pre = pre.right;
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}
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pre.right = root.right;
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root.right = root.left;
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root.left = null;
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}
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root = root.right;
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}
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}
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```
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### **C++**
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```cpp
@@ -180,6 +144,31 @@ public:
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### **Go**
182146
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```go
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/**
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* Definition for a binary tree node.
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* type TreeNode struct {
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* Val int
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* Left *TreeNode
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* Right *TreeNode
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* }
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*/
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func flatten(root *TreeNode) {
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for root != nil {
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if root.Left != nil {
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pre := root.Left
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for pre.Right != nil {
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pre = pre.Right
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}
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pre.Right = root.Right
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root.Right = root.Left
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root.Left = nil
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}
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root = root.Right
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}
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}
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```
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```go
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/**
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* Definition for a binary tree node.
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}
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```
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### **TypeScript**
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```ts
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/**
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* Definition for a binary tree node.
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* class TreeNode {
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* val: number
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* left: TreeNode | null
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* right: TreeNode | null
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* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
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* this.val = (val===undefined ? 0 : val)
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* this.left = (left===undefined ? null : left)
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* this.right = (right===undefined ? null : right)
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* }
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* }
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*/
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/**
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Do not return anything, modify root in-place instead.
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*/
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function flatten(root: TreeNode | null): void {
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while (root != null) {
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if (root.left != null) {
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let pre = root.left;
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while (pre.right != null) {
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pre = pre.right;
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}
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pre.right = root.right;
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root.right = root.left;
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root.left = null;
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}
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root = root.right;
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}
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}
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```
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### **JavaScript**
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```js
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/**
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* Definition for a binary tree node.
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* function TreeNode(val, left, right) {
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* this.val = (val===undefined ? 0 : val)
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* this.left = (left===undefined ? null : left)
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* this.right = (right===undefined ? null : right)
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* }
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*/
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/**
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* @param {TreeNode} root
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* @return {void} Do not return anything, modify root in-place instead.
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*/
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var flatten = function (root) {
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while (root) {
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if (root.left) {
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let pre = root.left;
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while (pre.right) {
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pre = pre.right;
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}
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pre.right = root.right;
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root.right = root.left;
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root.left = null;
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}
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root = root.right;
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}
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};
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```
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### **...**
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```

‎solution/0100-0199/0114.Flatten Binary Tree to Linked List/Solution.js

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@@ -0,0 +1,26 @@
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/**
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* Definition for a binary tree node.
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* function TreeNode(val, left, right) {
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* this.val = (val===undefined ? 0 : val)
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* this.left = (left===undefined ? null : left)
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* this.right = (right===undefined ? null : right)
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* }
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*/
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/**
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* @param {TreeNode} root
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* @return {void} Do not return anything, modify root in-place instead.
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*/
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var flatten = function (root) {
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while (root) {
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if (root.left) {
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let pre = root.left;
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while (pre.right) {
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pre = pre.right;
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}
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pre.right = root.right;
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root.right = root.left;
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root.left = null;
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}
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root = root.right;
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}
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};

‎solution/0100-0199/0114.Flatten Binary Tree to Linked List/Solution.py

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@@ -5,7 +5,7 @@
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# self.left = left
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# self.right = right
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class Solution:
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def flatten(self, root: TreeNode) -> None:
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def flatten(self, root: Optional[TreeNode]) -> None:
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"""
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Do not return anything, modify root in-place instead.
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"""

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