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0145.Binary Tree Postorder Traversal

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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。

Python3

# 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

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> 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);
        }
    }
}

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