108. Convert Sorted Array to Binary Search Tree

problem description

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

algorithm thought

这里需要生成一个平衡BST。平衡BST有很多种表示方法，但是我们只需要生成一种就行，那当然是生成最简单的bst。每次到一个节点时，找到中点，将一个有序数组对分。直到叶节点。这样就能保证最后得到的是平衡二叉树

code

/**
 * 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* sortedArrayToBST(vector<int>& nums) {
        return build(nums,0,nums.size());
    }
    TreeNode* build(vector<int>& nums,int low,int high){
        if(high==low)
            return NULL;
        int mid=low+((high-low)>>1);
        //cout<<low<<'-'<<high<<' ';
        TreeNode* root=new TreeNode(nums[mid]);
        root->left=build(nums,low,mid);
        root->right=build(nums,mid+1,high);
        return root;
    }
};

algorithm analysis

时间复杂度应该是O(n)遍历到所有节点，最后生成一个树