Path Sum算法详解

算法题目：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,

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

大致意思：给定一颗二叉树，决定是否存在一条从根结点到叶子结点的路径，使路径上的每个结点值的总和等于给定的sum，存在返回true，否则返回false。注意：一定是从根结点到叶子结点！！！

思路：递归做法，每到一个结点，用sum减去该结点的值，然后判断该结点是否为叶子结点，若是，判断sum==0?并返回true或false，否则就继续递归下去，递归退出条件为：树为空，返回false

    bool hasPathCore(TreeNode* root,int sum)    {        if(root==NULL)return false;        sum-=root->val;        if(root->left==NULL&&root->right==NULL)        {            if(sum==0)return true;            else return false;        }        if(hasPathCore(root->left,sum)||hasPathCore(root->right,sum))            return true;        else            return false;    }    bool hasPathSum(TreeNode* root, int sum) {        if(root==NULL)return false;        return hasPathCore(root,sum);    }