letecode [108] - Convert Sorted Array to Binary Search Tree

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

 

题目大意：

　　给定升序排序的数组，将其转换为一个平衡搜索二叉树。

理  解：

　　根据搜索二叉树的定义，该数组为二叉树的中序遍历结果。又该二叉树为平衡二叉树。输出可能二叉树结果。

　　用递归的思想，以数组中间的值为根节点，左右序列又可看作左右子树的数组，递归至叶节点。

　　从而构建二叉树，同时，也保证了构建的二叉树既是平衡二叉树又是搜索二叉树。

代 码 C++：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* AnSortedArrayToBST(vector<int>& nums,int left,int right) {
        if(left > right) return NULL;
        int mid = (right-left)/2+left;
        TreeNode* node = new TreeNode(nums[mid]);
        node->left = AnSortedArrayToBST(nums,left,mid-1);
        node->right = AnSortedArrayToBST(nums,mid+1,right);
        return node;
    }
    
    TreeNode* sortedArrayToBST(vector<int>& nums) {
        int n = nums.size();
        if(n==0) return NULL;
        return AnSortedArrayToBST(nums,0,n-1);
    }
};

运行结果：

　　执行用时 : 28 ms　　内存消耗 : 21.2 MB