108. Convert Sorted Array to Binary Search Tree

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于 2016-03-14 08:16:00 发布

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原文链接：http://www.cnblogs.com/neweracoding/p/5274485.html

本文介绍了一种将有序数组转换为高度平衡的二叉搜索树的方法。通过递归方式选择中间元素作为根节点，并将左右两侧的子数组分别递归转化为左子树和右子树。

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

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(M) Convert Sorted List to Binary Search Tree

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode sortedArrayToBST(int[] nums) {
        return createTree(nums, 0, nums.length);
    }
    
    private TreeNode createTree(int[] nums, int from, int to)
    {
        if(from>=to)
            return null;
        int len = to - from;
        if(len==1)
            return new TreeNode(nums[from]);
        
        int rootIndexOffset = len%2==0 ? len/2 : (len-1)/2;
        int rootIndex = from + rootIndexOffset;
        TreeNode root = new TreeNode(nums[rootIndex]);
        root.left = createTree(nums, from, rootIndex);
        root.right = createTree(nums, rootIndex+1, to);

        return root;
    }
}