145. Binary Tree Postorder Traversal

Problem Link

Description

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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]

Explanation:

Example 2:

Input: root = [1,2,3,4,5,null,8,null,null,6,7,9]

Output: [4,6,7,5,2,9,8,3,1]

Explanation:

Example 3:

Input: root = []

Output: []

Example 4:

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

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Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def postorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        res = []
 
        def go(node):
            if not node: return
 
            go(node.left)
            go(node.right)
            res.append(node.val)
        
        go(root)
        return res

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 {
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;
    }
};