feat: add solutions to lc problem: No.0094.Binary Tree Inorder Traversal

Author: yanglbme

solution/0000-0099/0094.Binary Tree Inorder Traversal/README.md

@@ -58,12 +58,15 @@
 
 <p><strong>进阶:</strong> 递归算法很简单，你可以通过迭代算法完成吗？</p>
 
-
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->
 
-递归遍历或利用栈实现非递归遍历。
+**1. 递归遍历**
+
+先递归左子树，再访问根节点，接着递归右子树。
+
+**2. 栈实现非递归遍历**
 
 非递归的思路如下：
 
@@ -72,6 +75,18 @@
 3. 左节点为空时，弹出栈顶元素并处理
 4. 重复 2-3 的操作
 
+**3. Morris 实现前序遍历**
+
+Morris 遍历无需使用栈，空间复杂度为 O(1)。核心思想是：
+
+遍历二叉树节点，
+
+1. 若当前节点 root 的左子树为空，**将当前节点值添加至结果列表 res** 中，并将当前节点更新为 `root.right`
+2. 若当前节点 root 的左子树不为空，找到左子树的最右节点 pre（也即是 root 节点在中序遍历下的前驱节点）：
+   - 若前驱节点 pre 的右子树为空，将前驱节点的右子树指向当前节点 root，并将当前节点更新为 `root.left`。
+   - 若前驱节点 pre 的右子树不为空，**将当前节点值添加至结果列表 res** 中，然后将前驱节点右子树指向空（即解除 pre 与 root 的指向关系），并将当前节点更新为 `root.right`。
+3. 循环以上步骤，直至二叉树节点为空，遍历结束。
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -89,17 +104,19 @@
 #         self.right = right
 class Solution:
     def inorderTraversal(self, root: TreeNode) -> List[int]:
+        res = []
+
         def inorder(root):
             if root:
                 inorder(root.left)
                 res.append(root.val)
                 inorder(root.right)
-        res = []
+
         inorder(root)
         return res
 ```
 
-非递归：
+栈实现非递归：
 
 ```python
 # Definition for a binary tree node.
@@ -110,8 +127,7 @@ class Solution:
 #         self.right = right
 class Solution:
     def inorderTraversal(self, root: TreeNode) -> List[int]:
-        s = []
-        res = []
+        res, s = [], []
         while root or s:
             if root:
                 s.append(root)
@@ -123,6 +139,36 @@ class Solution:
         return res
 ```
 
+Morris 遍历:
+
+```python
+# 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 inorderTraversal(self, root: TreeNode) -> List[int]:
+        res = []
+        while root:
+            if root.left is None:
+                res.append(root.val)
+                root = root.right
+            else:
+                pre = root.left
+                while pre.right and pre.right != root:
+                    pre = pre.right
+                if pre.right is None:
+                    pre.right = root
+                    root = root.left
+                else:
+                    res.append(root.val)
+                    pre.right = None
+                    root = root.right
+        return res
+```
+
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
@@ -146,26 +192,23 @@ class Solution:
  * }
  */
 class Solution {
-
-    private List<Integer> res;
-
     public List<Integer> inorderTraversal(TreeNode root) {
-        res = new ArrayList<>();
-        inorder(root);
+        List<Integer> res = new ArrayList<>();
+        inorder(root, res);
         return res;
     }
 
-    private void inorder(TreeNode root) {
+    private void inorder(TreeNode root, List<Integer> res) {
         if (root != null) {
-            inorder(root.left);
+            inorder(root.left, res);
             res.add(root.val);
-            inorder(root.right);
+            inorder(root.right, res);
         }
     }
 }
 ```
 
-非递归：
+栈实现非递归：
 
 ```java
 /**
@@ -185,63 +228,215 @@ class Solution {
  */
 class Solution {
     public List<Integer> inorderTraversal(TreeNode root) {
-        if (root == null) {
-            return Collections.emptyList();
-        }
         List<Integer> res = new ArrayList<>();
-        Deque<TreeNode> s = new ArrayDeque<>();
+        Deque<TreeNode> s = new LinkedList<>();
         while (root != null || !s.isEmpty()) {
             if (root != null) {
-                s.push(root);
+                s.offerLast(root);
                 root = root.left;
             } else {
-                root = s.pop();
+                root = s.pollLast();
+                res.add(root.val);
+                root = root.right;
+            }
+        }
+        return res;
+    }
+}
+```
+
+Morris 遍历:
+
+```java
+/**
+ * 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 List<Integer> inorderTraversal(TreeNode root) {
+        List<Integer> res = new ArrayList<>();
+        while (root != null) {
+            if (root.left == null) {
                 res.add(root.val);
                 root = root.right;
+            } else {
+                TreeNode pre = root.left;
+                while (pre.right != null && pre.right != root) {
+                    pre = pre.right;
+                }
+                if (pre.right == null) {
+                    pre.right = root;
+                    root = root.left;
+                } else {
+                    res.add(root.val);
+                    pre.right = null;
+                    root = root.right;
+                }
             }
         }
         return res;
     }
 }
 ```
+
 ### **JavaScript**
 
 递归：
 
 ```js
-var inorderTraversal = function(root) {
-    let res = [];
-    function inorder(root){
-        if(root){
-        inorder(root.left);
-        res.push(root.val);
-        inorder(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 {number[]}
+ */
+var inorderTraversal = function (root) {
+  let res = [];
+  function inorder(root) {
+    if (root) {
+      inorder(root.left);
+      res.push(root.val);
+      inorder(root.right);
     }
-    inorder(root);
-    return res;
+  }
+  inorder(root);
+  return res;
 };
 ```
-非递归：
+
+栈实现非递归：
 
 ```js
+/**
+ * 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 {number[]}
+ */
 var inorderTraversal = function (root) {
-    let res = [], stk = [];
-    let cur = root;
-    while (cur || stk.length !== 0) {
-        while (cur) {
-            stk.push(cur);
-            cur = cur.left;
-        } 
-        let top = stk.pop();
-        res.push(top.val);
-        cur = top.right;
+  let res = [];
+  let s = [];
+  while (root || s.length > 0) {
+    if (root) {
+      s.push(root);
+      root = root.left;
+    } else {
+      root = s.pop();
+      res.push(root.val);
+      root = root.right;
+    }
+  }
+  return res;
+};
+```
 
+### **C++**
+
+```cpp
+/**
+ * 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:
+    vector<int> inorderTraversal(TreeNode* root) {
+        vector<int> res;
+        while (root)
+        {
+            if (root->left == nullptr)
+            {
+                res.push_back(root->val);
+                root = root->right;
+            } else {
+                TreeNode* pre = root->left;
+                while (pre->right && pre->right != root)
+                {
+                    pre = pre->right;
+                }
+                if (pre->right == nullptr)
+                {
+                    pre->right = root;
+                    root = root->left;
+                }
+                else
+                {
+                    res.push_back(root->val);
+                    pre->right = nullptr;
+                    root = root->right;
+                }
+            }
+        }
+        return res;
     }
-    return res;
 };
 ```
 
+### **Go**
+
+```go
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func inorderTraversal(root *TreeNode) []int {
+	var res []int
+	for root != nil {
+		if root.Left == nil {
+			res = append(res, root.Val)
+			root = root.Right
+		} else {
+			pre := root.Left
+			for pre.Right != nil && pre.Right != root {
+				pre = pre.Right
+			}
+			if pre.Right == nil {
+				pre.Right = root
+				root = root.Left
+			} else {
+				res = append(res, root.Val)
+				pre.Right = nil
+				root = root.Right
+			}
+		}
+	}
+	return res
+}
+```
+
 ### **...**
 
 ```