学步园

题目：

给定一个有序的数组，要求转换为BST

思路：

递归。

每次找到数组中值作为BST的root，然后递归处理左半数组和右半数组分别作为左右子树

package Level2;

import Utility.TreeNode;

/**
 * Convert Sorted Array to Binary Search Tree  
 * 
 * Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
 */
public class S108 {

	public static void main(String[] args) {
		int[] num = {1,3};
		TreeNode n = sortedArrayToBST(num);
		n.print();
	}
	
	public static TreeNode sortedArrayToBST(int[] num) {
		return sortedArrayToBSTRec(num, 0, num.length-1);
    }
	
	// 二分查找节点值，并递归创建节点
	private static TreeNode sortedArrayToBSTRec(int[] num, int low, int high){
		if(low > high){
			return null;
		}
		int mid = low + (high-low)/2;		// 找到中值
		TreeNode root = new TreeNode(num[mid]);		// 创建根节点
        root.left = sortedArrayToBSTRec(num, low, mid-1);		// 递归创建左子树和右子树
        root.right = sortedArrayToBSTRec(num, mid+1, high);
		return root;
	}

}

/**
 * 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) {
        return rec(num, 0, num.length-1);
    }
    
    public TreeNode rec(int[] num, int low, int high){
        if(low > high){
            return null;
        }
        int mid = low + (high-low)/2;
        TreeNode root = new TreeNode(num[mid]);
        root.left = rec(num, low, mid-1);
        root.right = rec(num, mid+1, high);
        return root;
    }
}