LEETCODE--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.
方法一：
利用RedFlag函数将tree进行判断；
将flag作为一个标记；
如果满足条件则变为1；

/** * 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:    void RedFlag(TreeNode* root, int sum, int val, bool &flag){        val += root->val;        if(!root->left && !root->right && val== sum)            flag = 1;        if(root->left)            RedFlag(root->left, sum, val, flag);        if(root->right)            RedFlag(root->right, sum, val, flag);    }    bool hasPathSum(TreeNode* root, int sum) {        int val = 0;        bool flag = 0;        if(!root)            return 0;        RedFlag(root, sum, val, flag);        return flag;    }};

方法二：

/** * 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 YesOrNo(TreeNode* root, int sum, int val){        if(!root)            return 0;        val += root->val;        if(!root->left && !root->right){            if(val == sum)                return 1;            else                return 0;        }        return YesOrNo(root->left, sum, val) || YesOrNo(root->right, sum, val);     }    bool hasPathSum(TreeNode* root, int sum) {        int val = 0;        return YesOrNo(root, sum , val);    }};