145. Binary Tree Postorder Traversal

Description of Problem

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]

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?

Solution (Failed to answer the follow-up)

Code (C++)

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
private:
    void helper(TreeNode * root, vector<int> &vec){
        if(root == nullptr) 
            return;
        helper(root->left, vec);
        helper(root->right, vec);
        vec.push_back(root->val);
    }

public:
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> vec;
        helper(root, vec);
        return vec;
    }
};