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0112._Path_Sum.md

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112. Path Sum

难度:Easy

刷题内容

原题连接

内容描述

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.

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.

> 思路
******- 时间复杂度: O(N)******- 空间复杂度: O(1)******
DFS遍历二叉树,分别计算左子树和右子树的是否存在 sum - val 的路径

```cpp
/**
 * 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 == nullptr)
            return false;
        if(root ->right == nullptr && root ->left == nullptr)
            return sum - root ->val ? false : true;
        return hasPathSum(root ->left,sum - root ->val) || hasPathSum(root ->right,sum - root ->val); 
    }
};