需要另外设两个functions。 traverseTree function - preOrder 遍历每一个点,然后把每一个点代入到findPathSum function。 findPathSum function - 找到从root 开始到各个点的 path sum, 和sum比较,一样的话,就res++。 利用traverseTree 来把每一个点(不同的level)代入findPathSum, 来找到从那个点开始往下的各...
}publicvoidhelper(TreeNode root,inttarget,intcurrSum, Map<Integer, Integer>preSum) {if(root ==null) {return; } currSum+=root.val;if(preSum.containsKey(currSum -target)) { count+= preSum.get(currSum -target); }if(preSum.containsKey(currSum)) { preSum.put(currSum, preSum.get(currSum)...
You are given a binary tree in which each node contains an integer value. Find the number of paths that sum to a given value. The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes). The tree has ...
题目地址:https://leetcode.com/problems/path-sum/#/description 题目描述 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. ...
LeetCode 112 Path Sum 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. Note: A leaf is a node with no children. Example: Given the below binary tree and sum = 22...
1. Description Path Sum III 2. Solution Recursive /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
path(root.left,sum); path(root.right,sum); } } 113. 路径总和 II 给定一个二叉树和一个目标和,找到所有从根节点到叶子节点路径总和等于给定目标和的路径。 说明:叶子节点是指没有子节点的节点。 示例: 给定如下二叉树,以及目标和sum = 22,
publicintMaxPathSum(TreeNode root) { MaxPath(root); return_max; } privateintMaxPath(TreeNode current) { if(current == null) return0; // 如果子树路径和为负则应当置0表示最大路径不包含子树 intleft = Math.Max(MaxPath(current.left),0); ...
【CSON】LeetCode讲解 64. Minimum Path Sum发布于 2022-01-13 11:41 · 274 次播放 赞同添加评论 分享收藏喜欢 举报 力扣(LeetCode)算法编程程序员面试移动社交 写下你的评论... 还没有评论,发表第一个评论吧相关推荐 10:51 大型MMORPG「我们结婚吧」,玩家流失严重 #零基础看懂...
Left, remainSum) { return true } // 当右子结点存在,且存在一条右子结点到叶子路径上所有值到和为 remainSum ,则满足题意 if root.Right != nil && hasPathSum(root.Right, remainSum) { return true } // 左右子结点都不满足题意,返回 false return false } 题目链接: Path Sum: leetcode.com...