Lintcode177-Convert Sorted Array to Binary Search Tree With Minimal Height-Easy

Posted 2021-02-15 jessiezyr

177. Convert Sorted Array to Binary Search Tree With Minimal Height

Given a sorted (increasing order) array, Convert it to create a binary tree with minimal height.

Example

Example 1:
	Input: {1,2}
	Output:  A binary search tree with minimal height.
	
	Explanation:
	  2
	 /
	1


Example 2:
	Input: {1,2,3,4,5,6,7}
	Output:  A binary search tree with minimal height.
	
	Explanation:

         4
       /         2     6
     /     /     1   3  5   7

Notice

There may exist multiple valid solutions, return any of them.

 

思路：

Binary Search Tree特性：Binary Search Tree中序遍历的结果是升序数组。对于一个升序数组，一旦确定了根节点，根节点左半边部分全部属于左子树，根节点右半部分全部属于右子树。

最小高度的二叉树，就要尽可能满足其平衡。也就是说尽量保证根节点的左子树和右子树的节点个数差不多。所以一开始把数组中间的那个数定为根节点。

一旦确定了根节点在数组中的index值，其左子树的范围则是A[start, index - 1]，右子树的范围则是A[index + 1, end], 然后用分治法递归，求出根的左子树和右子树

代码：

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */


public class Solution {
    /*
     * @param A: an integer array
     * @return: A tree node
     */
    public TreeNode sortedArrayToBST(int[] A) {
        
        if (A.length == 0 || A == null) {
            return null;
        }
        return helper(A, 0, A.length - 1);
    }
    public TreeNode helper(int[] A, int start, int end) {
        if (start > end) {
            return null;
        }
        if (start == end) {
            return new TreeNode(A[start]);
        }
        int mid = (start + end) / 2;
        TreeNode root = new TreeNode(A[mid]);
        root.left = helper(A, start, mid - 1);
        root.right = helper(A, mid + 1, end);
        return root;
    }
}