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README.md

145. 二叉树的后序遍历

English Version

题目描述

给定一个二叉树，返回它的 后序 遍历。

示例:

输入: [1,null,2,3]  
   1
    \
     2
    /
   3 

输出: [3,2,1]

进阶: 递归算法很简单，你可以通过迭代算法完成吗？

解法

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 res 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 res 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 res 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”。

Python3

递归：

# 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: TreeNode) -> List[int]:
        res = []

        def postorder(root):
            if root:
                postorder(root.left)
                postorder(root.right)
                res.append(root.val)

        postorder(root)
        return res

栈实现非递归：

# 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: TreeNode) -> List[int]:
        res = []
        if root:
            s = [root]
            while s:
                node = s.pop()
                res.append(node.val)
                if node.left:
                    s.append(node.left)
                if node.right:
                    s.append(node.right)
        return res[::-1]

Morris 遍历：

# 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: TreeNode) -> List[int]:
        res = []
        while root:
            if root.right is None:
                res.append(root.val)
                root = root.left
            else:
                next = root.right
                while next.left and next.left != root:
                    next = next.left
                if next.left is None:
                    res.append(root.val)
                    next.left = root
                    root = root.right
                else:
                    next.left = None
                    root = root.left
        return res[::-1]

Java

递归：

/**
 * 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) {
        List<Integer> res = new ArrayList<>();
        postorder(root, res);
        return res;
    }

    private void postorder(TreeNode root, List<Integer> res) {
        if (root != null) {
            postorder(root.left, res);
            postorder(root.right, res);
            res.add(root.val);
        }
    }
}

栈实现非递归：

/**
 * 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> res = new LinkedList<>();
        if (root != null) {
            Deque<TreeNode> s = new LinkedList<>();
            s.offerLast(root);
            while (!s.isEmpty()) {
                TreeNode node = s.pollLast();
                res.addFirst(node.val);
                if (node.left != null) {
                    s.offerLast(node.left);
                }
                if (node.right != null) {
                    s.offerLast(node.right);
                }
            }
        }
        return res;
    }
}

Morris 遍历：

/**
 * 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> res = new LinkedList<>();
        while (root != null) {
            if (root.right == null) {
                res.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) {
                    res.addFirst(root.val);
                    next.left = root;
                    root = root.right;
                } else {
                    next.left = null;
                    root = root.left;
                }
            }
        }
        return res;
    }
}

C++

/**
 * 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> res;
        while (root)
        {
            if (root->right == nullptr)
            {
                res.push_back(root->val);
                root = root->left;
            }
            else
            {
                TreeNode *next = root->right;
                while (next->left && next->left != root)
                {
                    next = next->left;
                }
                if (next->left == nullptr)
                {
                    res.push_back(root->val);
                    next->left = root;
                    root = root->right;
                }
                else
                {
                    next->left = nullptr;
                    root = root->left;
                }
            }
        }
        reverse(res.begin(), res.end());
        return res;
    }
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func postorderTraversal(root *TreeNode) []int {
	var res []int
	for root != nil {
		if root.Right == nil {
			res = append([]int{root.Val}, res...)
			root = root.Left
		} else {
			next := root.Right
			for next.Left != nil && next.Left != root {
				next = next.Left
			}
			if next.Left == nil {
				res = append([]int{root.Val}, res...)
				next.Left = root
				root = root.Right
			} else {
				next.Left = nil
				root = root.Left
			}
		}
	}
	return res
}

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