【leetcode】path sum--easy

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.

//二叉树为空时，全部返回false

//从根结点到叶子结点的和为sum，而不是到某一个结点处和为sum就行

//二叉树的值可以为负值

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;    

        else

        {

            int v=root->val;

            if(root->right==NULL&&root->left==NULL&&sum-v==0)

                return true;

            else if(root->left!=NULL&&hasPathSum(root->left,sum-v)==true)

                return true;

            else if(root->right!=NULL&&hasPathSum(root->right,sum-v)==true)

                return true;

            else

                return false;

        }

    }

};