力扣654. 最大二叉树

Problem: 654. 最大二叉树

题目描述

思路

对于构造二叉树这类问题一般都是利用先、中、后序遍历，再将原始问题分解得出结果

1.定义递归函数build，每次将一个数组中的最大值作为当前子树的根节点构造二叉树；
2.每次找取当前范围内的最大值，作为当前的根节点；
3.递归求取出其左子树与右子树

复杂度

时间复杂度:

$O(n^2)$；其中n为二叉树节点的个数

空间复杂度:

$O(n)$

Code

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {/*** Maximum Binary Tree** @param nums Given array* @return TreeNode*/public TreeNode constructMaximumBinaryTree(int[] nums) {return build(nums, 0, nums.length - 1);}/*** Construction of binary tree function implementation** @param nums Given array* @param low  Given the left endpoint of the array* @param high Given the right endpoint of the array* @return TreeNode*/TreeNode build(int[] nums, int low, int high) {if (low > high) {return null;}int index = -1;int maxVal = Integer.MIN_VALUE;for (int i = low; i <= high; ++i) {if (maxVal < nums[i]) {maxVal = nums[i];index = i;}}//The root node is constructed first,// and then the left and right subtrees are constructedTreeNode root = new TreeNode(maxVal);root.left = build(nums, low, index - 1);root.right = build(nums, index + 1, high);return root;}
}