﻿ 112. Path Sum 二叉树路径的和

### 112. Path Sum 二叉树路径的和

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

For example:
Given the below binary tree and sum = 22,
5
/
4   8
/   /
11  13  4
/
7    2      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

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1. /**
2. * Definition for a binary tree node.
3. * public class TreeNode {
4. * public int val;
5. * public TreeNode left;
6. * public TreeNode right;
7. * public TreeNode(int x) { val = x; }
8. * }
9. */
10. public class Solution {
11. public bool HasPathSum(TreeNode root, int sum) {
12. return dfs(root, sum);
13. }
14. private bool dfs(TreeNode root, int sum) {
15. if (root == null)
16. return false;
17. if (root.left == null && root.right == null && sum == root.val)
18. return true;
19. return dfs(root.left, sum - root.val) || dfs(root.right, sum - root.val);
20. }
21. }