題目

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
 

簡單說就是把一個有序陣列轉為平衡二叉樹。常見的二叉查詢樹，有可能退化為連結串列，時間複雜度從logn將退化為O（n），所以實際應用中的AVL，紅黑樹都是平衡二叉樹

我的嘗試

其實是構造一個平衡二叉樹，無非是節點值已經指定，沒有思路處理這類問題，看了disscuss之後，按照高手的思路自己寫了一下，看看是否可以一點就通。

先梳理一下思路，二叉平衡時，首先滿足二叉樹的條件，左側小於根節點，右側大於根節點。因為陣列有序，其實我們尋找中間的節點作為根節點就可以。如何保證平衡呢？每次都是中間元素作為根節點，所有左右節點數相差絕對值不會大於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;
        
    }
    
}
 

執行之後stackoverflow錯誤。遞迴呼叫createTree的入參寫錯了，修改後如下：

/**
 * 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;
        
    }
    
}