Leetcode (:P) Ath Sum

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such this 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           /   /           /  4         /  \              7    2      1

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

ways to search for branches

/** * 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;/ /Current value            //If it is already a leaf node and is equal, return True    if (!root->left &&!root->right && nowsum = = sum)    return true;        BOOL Found_left = false;    BOOL Found_right = false;            if (root->left)//left dial hand tree is not empty, then search in the left subtree    found_left = Findpath (root->left, nowsum, sum);        if (!found_left && root->right)//If Zuozi not found, search in right subtree    found_right = Findpath (root->right, nowsum, sum );        return Found_left | | found_right;//returns True    }    bool Haspathsum (treenode* root, int sum) {        if (root = NULL)    return as long as it is found in one branch false;    Findpath (root, 0, sum);    };

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Leetcode (:P) Ath Sum