# 【LeetCode】 112. Path Sum

## Description
> 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.

> Note: A leaf is a node with no children.

> 給一個二元樹和一個總和值，判斷這棵樹有沒有一條從頭到葉的路徑總和剛好為總和值。
> 注意：一個葉節點沒有小孩。

## Example:

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.

## Solution

* 只要注意葉節點必須沒有左子樹和右子樹就好了，接下來就直接照一般DFS去跑所有路徑即可。

### 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:
    bool hasPathSum(TreeNode* root, int sum) {
        if(!root)return false;
        sum -= root->val;
        
        if(!root->left && !root->right)
        {
            if(sum==0)return true;
        }
        if(hasPathSum(root->left,sum))
            return true;
        if(hasPathSum(root->right,sum))
            return true;
        
        return false;
    }
};

###### tags: `LeetCode` `C++`