LeetCode OJ：Path Sum（路径之和）

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

要求求出是否有一条路径和与给出的值相等，注意中间节点与叶子节点的判断：

/**
 * 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 hasPathSum(TreeNode* root, int sum) {
        if(root == NULL) return false;
            return checkSum(root, sum);
    }
    
    bool checkSum(TreeNode* root, int sum)
    {
        if(root != NULL && sum == root->val && root->left == NULL && root->right == NULL){
            return true;//上面这个判断确实是叶子节点，值也同时满足
        }
        else if(root == NULL)
            return false;
        else
            return checkSum(root->left, sum - root->val) || checkSum(root->right, sum - root->val);    
    }
};

 java版本的如下，递归版本的没上面那么麻烦：

public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if(root == null)
            return false;
        if(root.left == null && root.right == null){
            return sum == root.val;
        }
        return hasPathSum(root.left, sum - root.val) || 
               hasPathSum(root.right, sum - root.val);
    }
}