茂名营销型网站建设,东莞网站网络公司,怎么做论坛的网站,学编程的正规学校将有序数组转换为二叉搜索树
https://leetcode.cn/problems/convert-sorted-array-to-binary-search-tree/description/
描述
给你一个整数数组 nums #xff0c;其中元素已经按 升序 排列请你将其转换为一棵 平衡 二叉搜索树
示例 1 输入#xff1a;nums [-10,-3,0,5,9…将有序数组转换为二叉搜索树
https://leetcode.cn/problems/convert-sorted-array-to-binary-search-tree/description/
描述
给你一个整数数组 nums 其中元素已经按 升序 排列请你将其转换为一棵 平衡 二叉搜索树
示例 1 输入nums [-10,-3,0,5,9]
输出[0,-3,9,-10,null,5]
解释[0,-10,5,null,-3,null,9] 也将被视为正确答案示例 2 输入nums [1,3]
输出[3,1]
解释[1,null,3] 和 [3,1] 都是高度平衡二叉搜索树。提示
1 nums.length 1 0 4 10^4 104- 1 0 4 10^4 104 nums[i] 1 0 4 10^4 104nums 按 严格递增 顺序排列
Typescript 版算法实现 1 ) 方案1: 中序遍历总是选择中间位置左边的数字作为根节点
/*** Definition for a binary tree node.* class TreeNode {* val: number* left: TreeNode | null* right: TreeNode | null* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {* this.val (valundefined ? 0 : val)* this.left (leftundefined ? null : left)* this.right (rightundefined ? null : right)* }* }*/function sortedArrayToBST(nums: number[]): TreeNode | null {return helper(nums, 0, nums.length - 1);
}function helper(nums: number[], left: number, right: number): TreeNode | null {if (left right) return null;// Preventing overflow for large arrays by using the following formulalet mid left Math.floor((right - left) / 2);let root new TreeNode(nums[mid]);root.left helper(nums, left, mid - 1);root.right helper(nums, mid 1, right);return root;
}2 ) 方案2: 中序遍历总是选择中间位置右边的数字作为根节点
/*** Definition for a binary tree node.* class TreeNode {* val: number* left: TreeNode | null* right: TreeNode | null* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {* this.val (valundefined ? 0 : val)* this.left (leftundefined ? null : left)* this.right (rightundefined ? null : right)* }* }*/function sortedArrayToBST(nums: number[]): TreeNode | null {return helper(nums, 0, nums.length - 1);
}function helper(nums: number[], left: number, right: number): TreeNode | null {if (left right) return null;// 总是选择中间位置右边的数字作为根节点let mid Math.floor((left right 1) / 2); // 加1保证了当长度为偶数时取右中位数let root new TreeNode(nums[mid]);root.left helper(nums, left, mid - 1);root.right helper(nums, mid 1, right);return root;
}3 ) 方案3: 中序遍历选择任意一个中间位置数字作为根节点
/*** Definition for a binary tree node.* class TreeNode {* val: number* left: TreeNode | null* right: TreeNode | null* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {* this.val (valundefined ? 0 : val)* this.left (leftundefined ? null : left)* this.right (rightundefined ? null : right)* }* }*/function sortedArrayToBST(nums: number[]): TreeNode | null {return helper(0, nums.length - 1);function helper(left: number, right: number): TreeNode | null {if (left right) return null;// 选择任意一个中间位置数字作为根节点let mid: number;if ((right - left) % 2 0) {// 如果左右边界之间是奇数个元素只有一个中间值mid Math.floor((left right) / 2);} else {// 如果左右边界之间是偶数个元素随机选择一个中间值mid left Math.floor((right - left Math.random()) / 2);}const root new TreeNode(nums[mid]);root.left helper(left, mid - 1);root.right helper(mid 1, right);return root;}
}4 ) 方案4: 简单版本
/*** Definition for a binary tree node.* class TreeNode {* val: number* left: TreeNode | null* right: TreeNode | null* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {* this.val (valundefined ? 0 : val)* this.left (leftundefined ? null : left)* this.right (rightundefined ? null : right)* }* }*/function sortedArrayToBST(nums: number[]): TreeNode | null {if(!nums.length) return null// 二叉搜索树的中序遍历就是升序列表// 数组中间的位置可以作为树的根节点const mid Math.floor(nums.length / 2)const root new TreeNode(nums[mid])root.left sortedArrayToBST(nums.slice(0,mid))root.right sortedArrayToBST(nums.slice(mid1))return root
}
