05 August 2008
Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.
Example:
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Given the sorted array: [-10,-3,0,5,9],
One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:
0
/ \
-3 9
/ /
-10 5
这道题是要将有序数组转为二叉搜索树，二叉搜索树始终满足左<=根<=右的特性，如果将二叉搜索树按中序遍历的话，得到的就是一个有序数组。那么反过来，根节点应该是有序数组的中间点，从中间点分开为左右两个有序数组，在分别找出其中间点作为原中间点的左右两个子节点，这不就是是二分查找法的核心思想么。所以这道题考的就是二分查找法。
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/**
* 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 sortedArrayToBST(nums, 0, nums.length - 1);
}
private TreeNode sortedArrayToBST(int[] nums, int left, int right) {//传入起点和终点
if (left > right) {
return null;
}
int mid = left + (right - left) / 2;
TreeNode current = new TreeNode(nums[mid]);
current.left = sortedArrayToBST(nums, left, mid - 1);
current.right = sortedArrayToBST(nums, mid + 1, right);
return current;
}
}
也可以不使用额外的递归函数，而是在原函数中完成递归，由于原函数的参数是一个数组，所以当把输入数组的中间数字取出来后，需要把所有两端的数组组成一个新的数组，并且分别调用递归函数，并且连到新创建的cur结点的左右子结点上面