108. Convert Sorted Array to Binary Search Tree

Posted 2020-09-29 いいえ敗者

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.

 

给定有序数组构造二叉搜索树。考虑到二叉搜索树的性质（中序遍历二叉搜索树可以得到一个排序好的数组）

构造的过程如下 根结点的左孩子结点则选取根结点左边区域的中间数，右孩子结点同理

/**
 * 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() - 1);
    }
    TreeNode* build(vector<int>& nums,int left, int right) {
        if (left > right) {
            return NULL;
        }
        int mid = (left + right) / 2;
        TreeNode* Node = new TreeNode(nums[mid]);
        Node->left = build(nums, left, mid - 1);
        Node->right = build(nums, mid + 1, right);
        return Node;
    }
};