国内哪些网站是php做的,培训优化,中企高呈建设网站,网站建设怎么下载代码题目链接#xff1a;https://leetcode.cn/problems/path-sum/description/
思路
递归#xff0c;先序遍历二叉树#xff0c;每遍历一个节点便减去当前存储值#xff08;targetSum targetSum - root.val#xff09;#xff1b;当到达某个节点等于targetSum (targetSum…题目链接https://leetcode.cn/problems/path-sum/description/
思路
递归先序遍历二叉树每遍历一个节点便减去当前存储值targetSum targetSum - root.val当到达某个节点等于targetSum (targetSum root.val)判断该节点是否为叶子节点(root.left null root.right null)如果是那么返回true如果该节点不满足targetSum那么递归遍历左子树和右子树(hasPathSum(root.left,targetSum - root.val) || hasPathSum(root.right,targetSum - root.val))任意一个返回true就成功。
代码实现
class Solution {public boolean hasPathSum(TreeNode root, int targetSum) {if(rootnull){ //空节点return false;}if(root.left null root.right null){ // 该节点为叶子节点return targetSum root.val; //相等则为true}return hasPathSum(root.left,targetSum - root.val) || hasPathSum(root.right,targetSum - root.val);}
}
