【Leetcode】Convert Sorted Array to Binary Search Tree

最新推荐文章于 2025-09-09 20:07:09 发布

转载最新推荐文章于 2025-09-09 20:07:09 发布 · 142 阅读

文章标签：

#leetcode #TreeNode #java

本文介绍如何将一个升序排列的数组转换为高度平衡的二叉搜索树。通过递归结合二分查找的方法，确保了生成的二叉树在左右子树的高度上尽可能接近。

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

Java:

1. 不多说，递归+二分法

水印人生： http://gongxuns.blogspot.com/2012/12/leetcodeconvert-sorted-array-to-binary.html

/**
 * 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) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if(num.length==0) return null;
        return sortedArrayToBST(num,0,num.length-1);
    }
     
    public TreeNode sortedArrayToBST(int[] num, int start, int end){
        if(start>end) return null;
        int mid=(start+end)/2;
        TreeNode res = new TreeNode(num[mid]);
        res.left=sortedArrayToBST(num,start,mid-1);
        res.right=sortedArrayToBST(num,mid+1,end);
        return res;
    }
}

2. http://blog.youkuaiyun.com/u013027996/article/details/21323449

public class Solution {  
    public TreeNode sortedArrayToBST(int[] num) {  
        if (num == null || num.length == 0) {  
            return null;  
        }  
        int len = num.length;  
        return createBSTree(num, 0, len - 1);  
    }  
  
    private TreeNode createBSTree(int[] num, int low, int high) {  
        if (low > high) {  
            return null;  
        }  
        if (low == high) {  
            return new TreeNode(num[low]);  
        }  
        int mid = (low + high) >> 1;  
        TreeNode root = new TreeNode(num[mid]);  
        root.left = createBSTree(num, low, mid - 1);  
        root.right = createBSTree(num, mid + 1, high);  
        return root;  
    }  
}