LeetCode 145. Binary Tree Postorder Traversal [Medium] [C++] 解題筆記

 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]

Example 2:

Input: root = []
Output: []

Example 3:

Input: root = [1]
Output: [1]

Example 4:

Input: root = [1,2]
Output: [2,1]

Example 5:

Input: root = [1,null,2]
Output: [2,1]

想法:

    這題和 144. Binary Tree Preorder Traversal 類似直接先用 recursive 解了，有空再練習 iterative 的方式。

完整程式碼:

class Solution {

public:

vector<int> postorder;

void traverse(TreeNode* root) {

if (root == NULL) { return; }

traverse(root->left);

traverse(root->right);

postorder.emplace_back(root->val);

}

vector<int> postorderTraversal(TreeNode* root) {

traverse(root);

return postorder;

}

};