［抄题］: Given an integer array with no duplicates. A maximum tree building on this array is defined as follow: The root is the maximum number in the array. The left subtree is the maximum tree constructed from left part subarray divided by the maximum…

654. Maximum Binary Tree 题目大意: 意思就是给你一组数,先选一个最大的作为根,这个数左边的数组作为左子树,右边的数组作为右子树,重复上一步. 读完就知道是递归了. 这个题真尼玛恶心,对JavaScript的友好度简直为0,用Js的可以直接放弃了. function TreeNode(val) { this.val = val this.left = this.right = null } var constructMaximumBinaryTree = function…

题目: Given an integer array with no duplicates. A maximum tree building on this array is defined as follow: The root is the maximum number in the array. The left subtree is the maximum tree constructed from left part subarray divided by the maximum nu…

题目如下: Given an integer array with no duplicates. A maximum tree building on this array is defined as follow: The root is the maximum number in the array. The left subtree is the maximum tree constructed from left part subarray divided by the maximum…

Given an integer array with no duplicates. A maximum tree building on this array is defined as follow: The root is the maximum number in the array. The left subtree is the maximum tree constructed from left part subarray divided by the maximum number…

网址:https://leetcode.com/problems/maximum-binary-tree/ 参考: https://leetcode.com/problems/maximum-binary-tree/discuss/106146/C%2B%2B-O(N)-solution 我自己的做法是依照题意的逻辑,逐一递归的做法. O(n2) /** * Definition for a binary tree node. * struct TreeNode { * int val; * T…

作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 递归 日期 题目地址:https://leetcode.com/problems/maximum-binary-tree/description/ 题目描述 Given an integer array with no duplicates. A maximum tree building on this array is defined as fol…

Given an integer array with no duplicates. A maximum tree building on this array is defined as follow: The root is the maximum number in the array. The left subtree is the maximum tree constructed from left part subarray divided by the maximum number…

用一个整型数组构建一个二叉树,根结点是数组中的最大值,左右子树分别是根结点的值在数组中左右两边的部分. 分析,这是二叉树中比较容易想到的问题了,直接使用递归就行了,代码如下: class Solution(object): def constructMaximumBinaryTree(self, nums): """ :type nums: List[int] :rtype: TreeNode """ if not nums: return No…

每次找到数组中的最大值,然后递归的构建左右树 public TreeNode constructMaximumBinaryTree(int[] nums) { if (nums.length==0) return null; return builder(nums,0,nums.length-1); } public TreeNode builder(int[] nums,int sta,int end) { /* 思路就是每次找到最大值,然后分为两个子数组递归构建左右树 */ if (sta>…