[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);
    }
};