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.

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/** * 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 __hasPathSum(TreeNode* root, int sum, int tempSum, bool& success){        if(!root->left && !root->right){            if(tempSum + root->val == sum){                success = true;            }        }                tempSum += root->val;                if(root->left){            __hasPathSum(root->left, sum, tempSum, success);        }        if(root->right){            __hasPathSum(root->right, sum, tempSum, success);        }            }            bool hasPathSum(TreeNode* root, int sum) {        bool res = false;        if(!root){            return res;        }        __hasPathSum(root, sum, 0, res);        return res;    }};

总是写的不够简洁。

bool hasPathSum(TreeNode *root, int sum) {        if (root == NULL) return false;        if (root->val == sum && root->left ==  NULL && root->right == NULL) return true;        return hasPathSum(root->left, sum-root->val) || hasPathSum(root->right, sum-root->val);    }