【Leetcode】Path Sum

题目：

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.

解题思路：

如果节点为空，返回false。如果节点不为空，则在路径和上加上当前节点的值，并判断是否是叶子节点，如果是叶子节点则判断当前的路径和是否等于条件所给的值。如果不是叶子节点，则递归的访问节点的左儿子和右儿子并对两者的结果取或。

代码1：

/**
 * 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) {
         return hasPathSum(root,sum,0);
    }
private:
    bool hasPathSum(TreeNode *root, int sum, int CurrSum){
        if(!root)return false;
        CurrSum+=root->val;
        if((!root->left)&&(!root->right))return CurrSum==sum;
        return hasPathSum(root->left,sum,CurrSum)||hasPathSum(root->right,sum,CurrSum);
    }
};

代码2：可以不另写函数，只需在访问节点时从sum值中减去节点值，并将差值作为新的sum值输入到下一层的递归函数中即可。

/**
 * 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)return false;
        if((!root->left)&&(!root->right))return root->val==sum;
        return hasPathSum(root->left,sum-root->val)||hasPathSum(root->right,sum-root->val);
    }
};