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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

递归遍历或利用栈实现非递归遍历。

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 s2 中，最后遍历结束，输出收集栈元素，即是后序遍历：左、右、头。

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]:
        def postorder(root):
            if root:
                postorder(root.left)
                postorder(root.right)
                res.append(root.val)
        res = []
        postorder(root)
        return res

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]:
        if not root:
            return []
        s1 = [root]
        s2 = []
        while s1:
            node = s1.pop()
            s2.append(node.val)
            if node.left:
                s1.append(node.left)
            if node.right:
                s1.append(node.right)
        return s2[::-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 {

    private List<Integer> res;

    public List<Integer> postorderTraversal(TreeNode root) {
        res = new ArrayList<>();
        postorder(root);
        return res;
    }

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

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) {
        if (root == null) {
            return Collections.emptyList();
        }
        Deque<TreeNode> s1 = new ArrayDeque<>();
        List<Integer> s2 = new ArrayList<>();
        s1.push(root);
        while (!s1.isEmpty()) {
            TreeNode node = s1.pop();
            s2.add(node.val);
            if (node.left != null) {
                s1.push(node.left);
            }
            if (node.right != null) {
                s1.push(node.right);
            }
        }
        Collections.reverse(s2);
        return s2;
    }
}

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