LeetCode Top Interview Questions 108. Convert Sorted Array to Binary Search Tree (Java版; Easy)

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LeetCode Top Interview Questions 108. Convert Sorted Array to Binary Search Tree (Java版; Easy)

题目描述

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example:

Given the sorted array: [-10,-3,0,5,9],

One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

      0
     / \
   -3   9
   /   /
 -10  5

第一次做; 核心: 为了让树平衡, 所以让数组的中点作为根节点,数组中点的左边也这样处理, 数组中点的右边也这样处理,所以自然地使用递归实现; 关键还是要明白搜索二叉树的定义:对于树上的任意一个节点node, 如果node左子树的值都比node.val小, node右子树的值都比node.val大, 那么该树就是搜索二叉树; 细节:要明白递归函数base case中的if(left>right)的发生条件, 比如上一轮L=0,R=1时此时M=0,那么调用root.left=core(nums,L,M-1)=core(nums,0,-1)就会出现left>right的情况

class Solution {
    public TreeNode sortedArrayToBST(int[] nums) {
        if(nums==null || nums.length==0)
            return null;
        //
        return core(nums, 0, nums.length-1);
    }
    //递归函数逻辑:将[left,right]的值构造成搜索二叉树, 为了实现平衡特性,让nums[mid]作为根节点的值, 根节点连接左子树[left,mid-1]的根节点,根节点连接右子树[mid+1,right]的根节点,最后返回根节点; 因为用到了左右子树的根节点信息, 所以自己也得返回根节点
    public TreeNode core(int[] nums, int left, int right){
        //base case
        if(left==right)
            return new TreeNode(nums[left]);
        //上一轮递归中L=0,R=1, 那么M-1=-1
        if(left>right)
            return null;
        //
        int mid = left + ((right-left)>>1);
        TreeNode root = new TreeNode(nums[mid]);
        root.left = core(nums, left, mid-1);
        root.right = core(nums, mid+1, right);
        return root;
    }
}