大纲
- 题目
- 地址
- 内容
- 解题
- 代码地址
题目
地址
https://leetcode.com/problems/binary-tree-postorder-traversal/description/
内容
Given the root of a binary tree, return the postorder traversal of its nodes’ values.
Example 1:
Input: root = [1,null,2,3]
Output: [3,2,1]
Explanation:
Example 2:
Input: root = [1,2,3,4,5,null,8,null,null,6,7,9]
Output: [4,6,7,5,2,9,8,3,1]
Explanation:
Example 3:
Input: root = []
Output: []
Example 4:
Input: root = [1]
Output: [1]
Constraints:
- The number of the nodes in the tree is in the range [0, 100].
- -100 <= Node.val <= 100
解题
这题就是考察二叉树的后序遍历：先左子树，再右子树，最后根节点。
class Solution {
public:vector<int> postorderTraversal(TreeNode* root) {vector<int> result;postorderTraversalDfs(root, result);return result; }private:void postorderTraversalDfs(TreeNode* root, vector<int>& result) {if (root == nullptr) return;postorderTraversalDfs(root->left, result);postorderTraversalDfs(root->right, result);result.push_back(root->val);}
};
代码地址
https://github.com/f304646673/leetcode/tree/main/145-Binary-Tree-Postorder-Traversal/cplusplus
