LeetCode | 145. Binary Tree Postorder Traversal

 

題目：

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]

 

Constraints:

• The number of the nodes in the tree is in the range [0, 100].
• -100 <= Node.val <= 100

 

Follow up:

Recursive solution is trivial, could you do it iteratively?

 

程式碼：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void backPostorder(TreeNode* root, vector<int>& res) {
		if(root == NULL)
			return;
		backPostorder(root->left, res);
		backPostorder(root->right, res);
		res.push_back(root->val);
		return;
	}
	vector<int> postorderTraversal(TreeNode* root) {
        vector<int> res;
		if(root == NULL)
			return res;
		backPostorder(root, res);
		return res;
    }
};