📝 Problem Details

💡 Explanation of Solution

dfs helper function to populate the result vector by reference through recursion 
recursively explore left and right children until we reach a leaf node 
then process the current node 

⌛ Complexity Analysis

Time Complexity: O(n)
Space Complexity: O(n) worst case, O(log n) best case 

💻 Implementation of Solution

/**
 * 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 {
public:
    void dfs(TreeNode* root, vector<int>& res) {
        if(root == nullptr) return;
        dfs(root->left, res);
        dfs(root->right, res);
        res.push_back(root->val);
    }
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> res;
        dfs(root, res);
        return res;
    }
};