112. 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

,

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：就是不斷遍歷樹，把值加起來，看最後如果加總值是目標值而且已經是樹葉了，就是得證

/**
* 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) {
return solve(root, 0, sum);
}

bool solve(TreeNode* root, int sum, int target) {
946a

if(root == NULL) {
return false;
}

sum += root->val;
if(sum == target && root->left == NULL && root->right == NULL) {
return true;
}

return solve(root->left, sum, target) || solve(root->right, sum, target);
}
};