[LeetCode]112. Path Sum(判断二叉树根到叶路径和是否等于sum)

112. Path Sum

原题链接
相似题目题解：113. Path Sum II && 437. Path Sum III
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.

题目大意：
给一个二叉树和一个数sum, 判断二叉树中是否存在一条从根节点到叶子节点的路径，路径中所有节点的和等于sum，返回true或者false

思路：

• 用递归，挺简单的，但是要注意二叉树为空时的特殊情况

代码如下：

C++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 == nullptr)            return false;        else if(root->left==nullptr && root->right==nullptr && root->val==sum)            return true;        else            return hasPathSum(root->left, sum-root->val) || hasPathSum(root->right, sum-root->val);    }};