leetcode Convert Sorted Array to Binary Search Tree 2.20 难度系数2-优快云博客

本文链接：https://blog.youkuaiyun.com/yiding_he/article/details/18812169

本文介绍了一种将有序数组转换为高度平衡二叉搜索树(BST)的方法。通过递归地选择中间元素作为根节点，并将左右两侧的子数组分别转换为左子树和右子树，确保了树的平衡性。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Question：

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

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
   public TreeNode sortedArrayToBST(int[] num) {
		if(num.length==0) return null;
        if(num.length==1) return new TreeNode(num[0]);
        
        int mid = num.length/2;
        TreeNode root = new TreeNode(num[mid]);
        
        int lenleft = mid;
        int lenright = num.length-mid-1;
        int[] left  = new int[lenleft];
        int[] right = new int[lenright];
        for(int i=0; i<lenleft; i++) left[i] = num[i];
        for(int i=mid+1; i<num.length; i++) right[i-mid-1] = num[i];
        
        root.left = sortedArrayToBST(left);
        root.right = sortedArrayToBST(right);
        
        return root;
	}

	
}

public class Solution {
    public TreeNode sortedArrayToBST(int[] num) {
        if(num.length==0)
            return null;
        int mid = num.length/2;
        TreeNode root = new TreeNode(num[mid]);
        int leftLen = mid;
        int rightLen = num.length-mid-1;
        int[] left = new int[leftLen];
        int[] right = new int[rightLen];
        System.arraycopy(num,0,left,0,leftLen);
        System.arraycopy(num,mid+1,right,0,rightLen);
        root.left = sortedArrayToBST(left);
        root.right = sortedArrayToBST(right);
        return root;
    }
}