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.

第一次写了一个AC的解，总觉得别扭

/**
 * Definition for binary tree
 * 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) {
        if (root==0)
            return false;
        return helper(root,sum);
    }
    
    bool helper(TreeNode* root, int sum){
        if (root==0){
            if (sum==0)
                return true;
            else
                return false;
        }
        int val=root->val;
        if (root->left && root->right)
            return (helper(root->left, sum-val)|| helper(root->right, sum-val));
        else if (!root->right&& root->left)
            return helper(root->left, sum-val);
        else if (!root->left && root->right)
            return helper(root->right,sum-val);
        return sum-val==0;
    }
};

看了下别人的，找到差距了。。。其实终止条件不应该放在root==0,而是应该放在leaf上。。。

the second version:

/**
 * Definition for binary tree
 * 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) {
        if (root==0)
            return false;
        if (!root->left && !root->right && root->val==sum)
            return true;
        return hasPathSum(root->left, sum-root->val)|| hasPathSum(root->right, sum-root->val);
    }
};