将有序数组转换为二叉搜索树

LeetCode108 解题思路

二叉树的中序遍历是升序序列，数组有序，能够确保是中序遍历，但中序遍历没法彻底肯定一棵二叉树。即结果不惟一。
每次以数组的中间数字做为二叉树的“根”结点，左右子树的元素分别位于“根”结点的两遍，依次递归结束

代码

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode sortedArrayToBST(int[] nums) {
        return helper(nums,0,nums.length-1);
    }
    public TreeNode helper(int[] nums,int left,int right){
        if(left>right){
            return null;
        }
        int mid=(left+right)/2;
        TreeNode root=new TreeNode(nums[mid]);
        root.left=helper(nums,left,mid-1);
        root.right=helper(nums,mid+1,right);
        return root;
    }
}