145. Binary Tree Postorder Traversal

Difficulty: Hard

Frequency: N/A

Given a binary tree, return the postorder traversal of its nodes’ values.

For example:
Given binary tree {1,#,2,3},

   1
    \
     2
    /
   3

return [3,2,1].

Note: Recursive solution is trivial, could you do it iteratively?

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Code:

/**
 * 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:
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> result;
        dfs(root, result);
        return result;
    }
    void dfs(TreeNode* node, vector<int>& result) {
        if (!node) return;
        dfs(node->left, result);
        dfs(node->right, result);
        result.push_back(node->val);
    }
};

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Another solution:

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Things to learn: