# 112.Path Sum

**112.Path Sum**

难度:Easy

> 给定一个二叉树和一个目标和，判断该树中是否存在根节点到叶子节点的路径，这条路径上所有节点值相加等于目标和。

说明: 叶子节点是指没有子节点的节点。

```
示例: 
给定如下二叉树，以及目标和 sum = 22，

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \      \
        7    2      1
返回 true, 因为存在目标和为 22 的根节点到叶子节点的路径 5->4->11->2。
```

可以使用递归或者迭代，这里用递归。每次递归时用sum减去当前节点的值，如果到了叶子节点，则判断sum是否为零。

```
/**
 * 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) return !sum;
        return hasPathSum(root->left,sum) || hasPathSum(root->right,sum);
        
    }
};
```

> 执行用时 :24 ms, 在所有 C++ 提交中击败了61.20%的用户\
> 内存消耗 :19.8 MB, 在所有 C++ 提交中击败了49.82%的用户


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