【leetcode】Path Sum

Path Sum

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,



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

思路：

由于要记录当前的和，因此需要单独建立一个函数，进行递归，不能借用本身来递归。

/**
* 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 path(TreeNode *root, int tempsum, int sum)
{
if(root==NULL) return 0;

tempsum+=root->val;
if(root->left==NULL && root->right==NULL && tempsum==sum) return 1;
if(root->left==NULL && root->right==NULL && tempsum!=sum) return 0;
return  path(root->left,tempsum, sum)||path(root->right,tempsum,sum);
}

bool hasPathSum(TreeNode* root, int sum) {
return  path(root, 0, sum);
}
};