108. Convert Sorted Array to Binary Search Tree

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最新推荐文章于 2022-02-19 22:56:38 发布

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分类专栏： leetcode-easy

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本文介绍了一种将已排序的数组转换为高度平衡的二叉搜索树的方法。平衡二叉搜索树定义为任意节点的两个子树深度差不超过1的二叉树。示例中给出了一个具体实现，通过递归方式选择中间元素作为根节点，确保了树的平衡性。

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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

注意看清题意,只要二叉平衡搜索树即可，不是一定要和例子中的树一模一样。

/**
 * 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* func(int left,int right,vector<int> &nums){
        if(left>right){
            return NULL;
        }
        int mid=(left+right)/2; 
        TreeNode* T=new TreeNode(nums[mid]);
        T->left=func(left,mid-1,nums);
        T->right=func(mid+1,right,nums);
        return T;       
    }
    
    TreeNode* sortedArrayToBST(vector<int>& nums) {
        return func(0,nums.size()-1,nums);
        
    }
};