112. 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.

分析

计算从根结点到叶结点之和是否等于sum，所以只需判断sum减去当前结点值为0时，当前结点是否是叶结点即可，若不是，则递归判断左子树或者右子树是否符合条件，当遍历到叶结点sum仍不符合条件时，返回false。

/** * 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;        if(sum-root->val==0&&root->left==NULL&&root->right==NULL)            return true;        else            return hasPathSum(root->left,sum-root->val)||hasPathSum(root->right,sum-root->val);    }};