# 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?

My solution:
Data structure:
Steps:
Complexity:
Runtime:
Space:
Test cases:
Corner cases:
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);
}
};
```

Another solution:
Data structure:
steps:
Complexity:
Runtime:
Space:
Code:

Things to learn:

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