给你二叉树的根结点 root ，请你将它展开为一个单链表：
- 展开后的单链表应该同样使用
TreeNode，其中right子指针指向链表中下一个结点，而左子指针始终为null。 - 展开后的单链表应该与二叉树 先序遍历 顺序相同。
示例 1：
输入：root = [1,2,5,3,4,null,6] 输出：[1,null,2,null,3,null,4,null,5,null,6]
示例 2：
输入：root = [] 输出：[]
示例 3：
输入：root = [0] 输出：[0]
提示：
- 树中结点数在范围
[0, 2000]内 -100 <= Node.val <= 100
进阶：你可以使用原地算法（O(1) 额外空间）展开这棵树吗？
方法一：寻找前驱节点
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。
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。
时间复杂度
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def flatten(self, root: Optional[TreeNode]) -> None:
"""
Do not return anything, modify root in-place instead.
"""
while root:
if root.left:
pre = root.left
while pre.right:
pre = pre.right
pre.right = root.right
root.right = root.left
root.left = None
root = root.right/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public void flatten(TreeNode root) {
while (root != null) {
if (root.left != null) {
// 找到当前节点左子树的最右节点
TreeNode pre = root.left;
while (pre.right != null) {
pre = pre.right;
}
// 将左子树的最右节点指向原来的右子树
pre.right = root.right;
// 将当前节点指向左子树
root.right = root.left;
root.left = null;
}
root = root.right;
}
}
}/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
void flatten(TreeNode* root) {
while (root) {
if (root->left) {
TreeNode* pre = root->left;
while (pre->right) {
pre = pre->right;
}
pre->right = root->right;
root->right = root->left;
root->left = nullptr;
}
root = root->right;
}
}
};// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
// pub val: i32,
// pub left: Option<Rc<RefCell<TreeNode>>>,
// pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
// #[inline]
// pub fn new(val: i32) -> Self {
// TreeNode {
// val,
// left: None,
// right: None
// }
// }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
#[allow(dead_code)]
pub fn flatten(root: &mut Option<Rc<RefCell<TreeNode>>>) {
if root.is_none() {
return;
}
let mut v: Vec<Option<Rc<RefCell<TreeNode>>>> = Vec::new();
// Initialize the vector
Self::pre_order_traverse(&mut v, root);
// Traverse the vector
let n = v.len();
for i in 0..n - 1 {
v[i].as_ref().unwrap().borrow_mut().left = None;
v[i].as_ref().unwrap().borrow_mut().right = v[i + 1].clone();
}
}
#[allow(dead_code)]
fn pre_order_traverse(v: &mut Vec<Option<Rc<RefCell<TreeNode>>>>, root: &Option<Rc<RefCell<TreeNode>>>) {
if root.is_none() {
return;
}
v.push(root.clone());
let left = root.as_ref().unwrap().borrow().left.clone();
let right = root.as_ref().unwrap().borrow().right.clone();
Self::pre_order_traverse(v, &left);
Self::pre_order_traverse(v, &right);
}
}/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func flatten(root *TreeNode) {
for root != nil {
if root.Left != nil {
pre := root.Left
for pre.Right != nil {
pre = pre.Right
}
pre.Right = root.Right
root.Right = root.Left
root.Left = nil
}
root = root.Right
}
}/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func flatten(root *TreeNode) {
for root != nil {
left, right := root.Left, root.Right
root.Left = nil
if left != nil {
root.Right = left
for left.Right != nil {
left = left.Right
}
left.Right = right
}
root = root.Right
}
}/**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
/**
Do not return anything, modify root in-place instead.
*/
function flatten(root: TreeNode | null): void {
while (root !== null) {
if (root.left !== null) {
let pre = root.left;
while (pre.right !== null) {
pre = pre.right;
}
pre.right = root.right;
root.right = root.left;
root.left = null;
}
root = root.right;
}
}/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {void} Do not return anything, modify root in-place instead.
*/
var flatten = function (root) {
while (root) {
if (root.left) {
let pre = root.left;
while (pre.right) {
pre = pre.right;
}
pre.right = root.right;
root.right = root.left;
root.left = null;
}
root = root.right;
}
};