Convert Sorted Array to Binary Search Tree

描述

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

分析

二分法。

代码

Java

// Convert Sorted Array to Binary Search Tree
// 二分法，时间复杂度O(n)，空间复杂度O(logn)
public class Solution {
    public TreeNode sortedArrayToBST (int[] nums) {
        return sortedArrayToBST(nums, 0, nums.length);
    }

    private static TreeNode sortedArrayToBST (int[] nums, int begin, int end) {
        int length = end - begin;
        if (length < 1) return null;  // 终止条件

        // 三方合并
        int mid = begin + length / 2;
        TreeNode root = new TreeNode (nums[mid]);
        root.left = sortedArrayToBST(nums, begin, mid);
        root.right = sortedArrayToBST(nums, mid + 1, end);

        return root;
    }
}

C++

// Convert Sorted Array to Binary Search Tree
// 二分法，时间复杂度O(n)，空间复杂度O(logn)
class Solution {
public:
    TreeNode* sortedArrayToBST (vector<int>& num) {
        return sortedArrayToBST(num.begin(), num.end());
    }

    template<typename RandomAccessIterator>
    TreeNode* sortedArrayToBST (RandomAccessIterator first,
            RandomAccessIterator last) {
        const auto length = distance(first, last);

        if (length <= 0) return nullptr;  // 终止条件

        // 三方合并
        auto mid = first + length / 2;
        TreeNode* root = new TreeNode (*mid);
        root->left = sortedArrayToBST(first, mid);
        root->right = sortedArrayToBST(mid + 1, last);

        return root;
    }
};

相关题目

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