题目
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:
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简单说就是把一个有序数组转为平衡二叉树。常见的二叉查找树，有可能退化为链表，时间复杂度从logn将退化为O（n），所以实际应用中的AVL，红黑树都是平衡二叉树
我的尝试
其实是构造一个平衡二叉树，无非是节点值已经指定，没有思路处理这类问题，看了disscuss之后，按照高手的思路自己写了一下，看看是否可以一点就通。
先梳理一下思路，二叉平衡时，首先满足二叉树的条件，左侧小于根节点，右侧大于根节点。因为数组有序，其实我们寻找中间的节点作为根节点就可以。如何保证平衡呢？每次都是中间元素作为根节点，所有左右节点数相差绝对值不会大于1.
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode sortedArrayToBST(int[] nums) {
if(nums.length==0) return null;
int p=0;
int q=nums.length;
return createTree(nums,p,q);
}
public TreeNode createTree(int[] nums,int p,int q){
if(p>q){
return null;
}
int mid=(p+q)/2;
TreeNode treeNode=new TreeNode(nums[mid]);
treeNode.left=createTree(nums,0,mid-1);
treeNode.right=createTree(nums,mid+1,nums.length);
return treeNode;
}
}运行之后stackoverflow错误。递归调用createTree的入参写错了，修改后如下：
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode sortedArrayToBST(int[] nums) {
if(nums.length==0) return null;
int p=0;
int q=nums.length-1;
return createTree(nums,p,q);
}
public TreeNode createTree(int[] nums,int p,int q){
if(p>q){
return null;
}
int mid=(p+q)/2;
TreeNode treeNode=new TreeNode(nums[mid]);
treeNode.left=createTree(nums,p,mid-1);
treeNode.right=createTree(nums,mid+1,q);
return treeNode;
}
}