classSolution {public:intdiameterOfBinaryTree(TreeNode*root) { diameterCore(root);returndiameter; }intdiameterCore(TreeNode*root){if(root ==NULL)return0;intleft = diameterCore(root->left);intright = diameterCore(root->right); diameter= max(diameter,left +right);returnleft > right ? left +...
int diameterOfBinaryTree(TreeNode* root) { diameterCore(root); return diameter; } int diameterCore(TreeNode* root){ if(root == NULL) return 0; int left = diameterCore(root->left); int right = diameterCore(root->right); diameter = max(diameter,left + right); return left > right ?
LeetCode 543. Diameter of Binary Tree 二叉树的直径 (C++/Java) LeetCode 687. Longest Univalue Path 最长同值路径 (C++/Java) 那么这道题还是从根节点递归求解,当前结点的最大路径,是当前节点的值加上左右孩子的最大路径,同时和全局的最大值进行比较,更新最大值,而作为返回值时,需要在左右孩子中选取最大...
一道难度为medium的题目,原题地址:https://leetcode.com/problems/maximum-binary-tree/description/,一道与树相关的题目,用递归的方法即可解决,执行时间为66ms。 题目: Given an integer array with no duplicates. A maximum ... LeetCode-Maximum Depth of N-ary Tree ...
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A path in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence at most once. Note that the path does not need to pass through the root. ...
543. Diameter of Binary Tree 问题: 求出给定二叉树中,最大路径(节点个数最多)。(至少包含树中一个节点) Example: Given a binary tree 1 / \ 2 3 / \ 4 5 Return 3,whichis the length of the path [4,2,1,3] or [5,2,1,3]. ...
124. Binary Tree Maximum Path Sum; 543. Diameter of Binary Tree; 687. Longest Univalue Path;求二叉树的最大路径,路径和,相同值节点的最大路径。 解法:Binary Tree (二叉树) 模版:一棵树问题: 1def solve(root)2//无效节点处理3ifnot root:return...4//递归终点,base5iff(root):return...6//左...
Leetcode 559. Maximum Depth of N-ary Tree c++,如果本节点为空,返回0,否则返回 这棵树孩子中(找到每个节点的最大值,返回最大值+1即可,1是本节点的深度) /*// Definition for a Node.class Node {public:int val;vector<Node*> children;Node() {}Node(int _val) {val = _val;}Node(int _val...