LeetCode（95） Unique Binary Search Trees II

题目

Given n, generate all structurally unique BST’s (binary search trees) that store values 1…n.

For example,
Given n = 3, your program should return all 5 unique BST’s shown below.

confused what “{1,#,2,3}” means? > read more on how binary tree is serialized on OJ.

OJ’s Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where ‘#’ signifies a path terminator where no node exists below.

Here’s an example:

The above binary tree is serialized as “{1,2,3,#,#,4,#,#,5}”.

分析

给定整数n，求输入元素为 [1，n] 时，所构成的全部二叉查找树；

我们都知道二叉查找树的特点，左子树节点值小于根节点，右子树节点值大于根节点。

对于输入 [1，n] ，每个值 i 都可以作为根节点，小于 i 的元素构成左子树，大于 i 的元素构成右子树。

所以，此题的解决办法为二叉树常用递归。

AC代码

class Solution {
public:
    vector<TreeNode*> generateTrees(int n) {
        if (n <= 0)
            return vector<TreeNode *>(1 , NULL);

        //对值为 [1 , n]的每个元素都可做二叉查找树的根节点
        return generateTrees(1, n);
    }

    //构造根节点[lhs , rhs]的所有二叉查找树
    vector<TreeNode *> generateTrees(int lhs, int rhs)
    {
        if (lhs > rhs)
        {
            return vector<TreeNode *>(1 , NULL);
        }

        //存储每个查找树的根节点
        vector<TreeNode *> ret;
        for (int r = lhs; r <= rhs; r++)
        {
            //[lhs~r-1]间节点作为左子树，[r+1~rhs]间节点作为右子树
            vector<TreeNode *> lefts = generateTrees(lhs, r - 1);
            vector<TreeNode *> rights = generateTrees(r + 1, rhs);

            //链接符合要求的左右子树
            int lsize = lefts.size();
            int rsize = rights.size();
            for (int i = 0; i < lsize; ++i)
            {
                for (int j = 0; j < rsize; ++j)
                {
                    //当前节点作为根节点
                    TreeNode *root = new TreeNode(r);
                    root->left = lefts[i];
                    root->right = rights[j];
                    ret.push_back(root);
                }//for
            }//for 
        }//for
        return ret;
    }
};

GitHub测试程序源码