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.

              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.

/**
 * 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 findPath(TreeNode* root, int nowSum,const int sum)
    {
    	nowSum += root->val;//当前值
    
        //假设已经是叶子结点，且和相等，则直接返回真
    	if (!root->left && !root->right && nowSum == sum)
    		return true;
    	
    	bool found_left = false;
    	bool found_right = false;
    
    
    	if (root->left)	//左子树不为空。则在左子树中搜索
    		found_left = findPath(root->left, nowSum, sum);
    	
    	if (!found_left && root->right)	//假设左子树没找到，则在右子树中搜索
    		found_right = findPath(root->right, nowSum, sum);	
    
    	return found_left || found_right;//仅仅要在一条支中上找到则返回真
    }

    bool hasPathSum(TreeNode* root, int sum) {
        if (root == NULL)
		    return false;
	    findPath(root, 0, sum);
    }
};