【Unique Binary Search Trees II】cpp

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.

   1         3     3      2      1
    \       /     /      / \           3     2     1      1   3      2
    /     /       \                    2     1         2                 3

代码:

/**
 * 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:
    vector<TreeNode*> generateTrees(int n) {
        return Solution::generateBST(1, n);
    }
        static vector<TreeNode*> generateBST(int min, int max)
        {
            vector<TreeNode*> ret;
            if ( min>max ) { ret.push_back(NULL); return ret; }
            for ( int i = min; i<=max; ++i )
            {
                vector<TreeNode*> left = Solution::generateBST(min, i-1);
                vector<TreeNode*> right = Solution::generateBST(i+1,max);
                for ( size_t l = 0; l < left.size(); ++l )
                {
                    for ( size_t r = 0; r < right.size(); ++r )
                    {
                        TreeNode *root = new TreeNode(i);
                        root->left = left[l];
                        root->right = right[r];
                        ret.push_back(root);
                    }
                }
            }
            return ret;
        }
};

tips:

直接学习大神的代码

http://bangbingsyb.blogspot.sg/2014/11/leetcode-unique-binary-search-trees-i-ii.html

一开始一直有一个疑问,如果min==max的时候(即只有一个元素的时候)能构造一个新的节点返回么?

肯定是可以的。因为这时left返回的是含有一个NULL的vector,right返回的是含有一个NULL的vector;两个vector的长度都是1,因此可以构造出这个新的点。

时间: 2024-10-25 05:51:17

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