题目要求:
给你二叉树的根节点 root 和一个表示目标和的整数 targetSum ,判断该树中是否存在 根节点到叶子节点 的路径,这条路径上所有节点值相加等于目标和 targetSum 。
叶子节点 是指没有子节点的节点。
示例 1:
输入:root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
输出:true
示例 2:输入:root = [1,2,3], targetSum = 5
输出:false
示例 3:输入:root = [1,2], targetSum = 0
输出:false
提示:树中节点的数目在范围 [0, 5000] 内
-1000 <= Node.val <= 1000
-1000 <= targetSum <= 1000来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/path-sum
解题代码:
//Definition for a binary tree node. struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode() : val(0), left(nullptr), right(nullptr) {} TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} }; class Solution { public: bool hasPathSum(TreeNode* root, int targetSum) { if (root == NULL) return false; if (root->left == NULL && root->right == NULL) return root->val == targetSum; return hasPathSum(root->left, targetSum - root->val) || hasPathSum(root->right, targetSum - root->val); } };