# [145. 二叉树的后序遍历](https://leetcode.cn/problems/binary-tree-postorder-traversal)
[English Version](/solution/0100-0199/0145.Binary%20Tree%20Postorder%20Traversal/README_EN.md)
## 题目描述
<!-- 这里写题目描述 -->
<p>给你一棵二叉树的根节点 <code>root</code> ，返回其节点值的 <strong>后序遍历 </strong>。</p>
<p> </p>
<p><strong>示例 1：</strong></p>
<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0145.Binary%20Tree%20Postorder%20Traversal/images/pre1.jpg" style="width: 127px; height: 200px;" />
<pre>
<strong>输入：</strong>root = [1,null,2,3]
<strong>输出：</strong>[3,2,1]
</pre>
<p><strong>示例 2：</strong></p>
<pre>
<strong>输入：</strong>root = []
<strong>输出：</strong>[]
</pre>
<p><strong>示例 3：</strong></p>
<pre>
<strong>输入：</strong>root = [1]
<strong>输出：</strong>[1]
</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li>树中节点的数目在范围 <code>[0, 100]</code> 内</li>
<li><code>-100 <= Node.val <= 100</code></li>
</ul>
<p> </p>
<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 -->
**1. 递归遍历**
先递归左右子树，再访问根节点。
**2. 栈实现非递归遍历**
非递归的思路如下：
先序遍历的顺序是：头、左、右，如果我们改变左右孩子的顺序，就能将顺序变成：头、右、左。
我们先不打印头节点，而是存放到另一个收集栈 ans 中，最后遍历结束，输出收集栈元素，即是后序遍历：左、右、头。收集栈是为了实现结果列表的逆序。我们也可以直接使用链表，每次插入元素时，放在头部，最后直接返回链表即可，无需进行逆序。
**3. Morris 实现后序遍历**
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 的右子树为空，**将当前节点值添加至结果列表 ans** 中，并将当前节点更新为 `root.left`
2. 若当前节点 root 的右子树不为空，找到右子树的最左节点 next（也即是 root 节点在中序遍历下的后继节点）：
- 若后继节点 next 的左子树为空，**将当前节点值添加至结果列表 ans** 中，然后将后继节点的左子树指向当前节点 root，并将当前节点更新为 `root.right`。
- 若后继节点 next 的左子树不为空，将后继节点左子树指向空（即解除 next 与 root 的指向关系），并将当前节点更新为 `root.left`。
3. 循环以上步骤，直至二叉树节点为空，遍历结束。
4. 最后返回结果列表的逆序即可。
> Morris 后序遍历跟 Morris 前序遍历思路一致，只是将前序的“根左右”变为“根右左”，最后逆序结果即可变成“左右根”。
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### **Python3**
<!-- 这里可写当前语言的特殊实现逻辑 -->
递归：
```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 postorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
def dfs(root):
if root is None:
return
dfs(root.left)
dfs(root.right)
nonlocal ans
ans.append(root.val)
ans = []
dfs(root)
return ans
```
栈实现非递归：
```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 postorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
ans = []
if root is None:
return ans
stk = [root]
while stk:
node = stk.pop()
ans.append(node.val)
if node.left:
stk.append(node.left)
if node.right:
stk.append(node.right)
return ans[::-1]
```
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 postorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
ans = []
while root:
if root.right is None:
ans.append(root.val)
root = root.left
else:
next = root.right
while next.left and next.left != root:
next = next.left
if next.left != root:
ans.append(root.val)
next.left = root
root = root.right
else:
next.left = None
root = root.left
return ans[::-1]
```
### **Java**
<!-- 这里可写当前语言的特殊实现逻辑 -->
递归：
```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 {
private List<Integer> ans;
public List<Integer> postorderTraversal(TreeNode root) {
ans = new ArrayList<>();
dfs(root);
return ans;
}
private void dfs(TreeNode root) {
if (root == null) {
return;
}
dfs(root.left);
dfs(root.right);
ans.add(root.val);
}
}
```
栈实现非递归：
```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> postorderTraversal(TreeNode root) {
LinkedList<Integer> ans = new LinkedList<>();
if (root == null) {
return ans;
}
Deque<TreeNode> stk = new ArrayDeque<>();
stk.push(root);
while (!stk.isEmpty()) {
TreeNode node = stk.pop();
ans.addFirst(node.val);
if (node.left != null) {
stk.push(node.left);
}
if (node.right != null) {
stk.push(node.right);
}
}
return ans;
}
}
```
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> postorderTraversal(TreeNode root) {
LinkedList<Integer> ans = new LinkedList<>();
while (root != null) {
if (root.right == null) {
ans.addFirst(root.val);
root = root.left;
} else {
TreeNode next = root.right;
while (next.left != null && next.left != root) {
next = next.left;
}
if (next.left == null) {
ans.addFirst(root.val);
next.left = root;
root = root.right;
} else {
next.left = null;
root = root.left;
}
}
}
return ans;
}
}
```
### **TypeScript**
```ts
/**
* 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)
* }
* }
*/
function postorderTraversal(root: TreeNode | null): number[] {
if (root == null) return [];
let stack = [];
let ans = [];
let prev = null;
while (root || stack.length) {
while (root) {
stack.push(root);
root = root.left;
}
root = stack.pop();
if (!root.right || root.right == prev) {
ans.push(root.val);
prev = root;
root = null;
} else {
stack.push(root);
root = root.right;
}
}
return ans;
}
```
```ts
/**
* 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)
* }
* }
*/
function postorderTraversal(root: TreeNode | null): number[] {
if (root == null) {
return [];
}
const { val, left, right } = root;
return [...postorderTraversal(left), ...postorderTraversal(right), val];
}
```
```ts
/**
* 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)
* }
* }
*/
function postorderTraversal(root: TreeNode | null): number[] {
const res = [];
while (root != null) {
const { val, left, right } = root;
if (right == null) {
res.push(val);
root = left;
} else {
let next = right;
while (next.left != null && next.left != root) {
next = next.left;
}
if (next.left == null) {
res.push(val);
next.left = root;
root = right;
} else {
next.left = null;
root = left;
}
}
}
return res.reverse();
}
```
### **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> postorderTraversal(TreeNode* root) {
vector<int> ans;
while (root)
{
if (!root->right)
{
ans.push_back(root->val);
root = root->left;
}
else
{
TreeNode* next = root->right;
while (next->left && next->left != root)
{
next = next->left;
}
if (!next->left)
{
ans.push_back(root->val);
next->left = root;
root = root->right;
}
else
{
next->left = nullptr;
root = root->left;
}
}
}
reverse(ans.begin(), ans.end());
return ans;
}
};
```
### **Go**
```go
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func postorderTraversal(root *TreeNode) []int {
var ans []int
for root != nil {
if root.Right == nil {
ans = append([]int{root.Val}, ans...)
root = root.Left
} else {
next := root.Right
for next.Left != nil && next.Left != root {
next = next.Left
}
if next.Left == nil {
ans = append([]int{root.Val}, ans...)
next.Left = root
root = root.Right
} else {
next.Left = nil
root = root.Left
}
}
}
return ans
}
```
### **Rust**
```rust
// 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 {
fn dfs(root: &Option<Rc<RefCell<TreeNode>>>, res: &mut Vec<i32>) {
if root.is_none() {
return;
}
let node = root.as_ref().unwrap().borrow();
Self::dfs(&node.left, res);
Self::dfs(&node.right, res);
res.push(node.val);
}
pub fn postorder_traversal(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
let mut res = vec![];
Self::dfs(&root, &mut res);
res
}
}
```
```rust
// 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 {
pub fn postorder_traversal(mut root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
let mut res = vec![];
let mut stack = vec![];
while root.is_some() || !stack.is_empty() {
if root.is_some() {
let next = root.as_mut().unwrap().borrow_mut().left.take();
stack.push(root);
root = next;
} else {
root = stack.pop().unwrap();
let next = root.as_mut().unwrap().borrow_mut().right.take();
if next.is_some() {
stack.push(root);
} else {
res.push(root.as_ref().unwrap().borrow().val);
}
root = next;
}
}
res
}
}
```
### **...**
```
```
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