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Leetcode P0108"Convert Sorted Array to Binary Search Tree" 题解

2020/09/07 Leetcode

博文中会简要介绍Leetcode P0108题目分析及解题思路。

“Convert Sorted Array to Binary Search Tree”是比较简单和基础的树的题目，运用了深度优先搜索，这里是递归回溯的思路。这道题需要我们构造一个平衡二叉搜索树，我们可以每次将整个数组二分，其中点作为BST的根结点，然后运用递归回溯的思路，向下层生长。

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.

以下是Java的题解代码实现。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode sortedArrayToBST(int[] nums) {
        if (nums == null || nums.length == 0)
            return null;
        
        return this.build(nums, 0, nums.length-1);
    }
    
    private TreeNode build(int[] nums, int left, int right) {
        if (left > right)
            return null;
        
        int mid = (left+right)/2;
        TreeNode root = new TreeNode(nums[mid]);
        
        root.left = this.build(nums, left, mid-1);
        root.right = this.build(nums, mid+1, right);
        
        return root;
    }
}
Java

以下是C++的题解代码实现。

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

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