LeetCode Unqiue Binary Search Trees 2

Description

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

The original problem is here.

The original code is here.

My Solution

I solve this problem in C++, as below:

/*
*Unique Binary Search Trees II 
*Author: shuaijiang
*Email: zhaoshuaijiang8@gmail.com
*/
#include<iostream>
#include<string.h>
#include<stdlib.h>
using namespace std;

/**
 * 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) {
        if(n == 0)
            return buildTree(1,0);
        return buildTree(1,n);
    }
    vector<TreeNode*> buildTree(int start, int end){
        vector<TreeNode*> tree;
        if(start > end){
            tree.push_back(NULL);
            return tree;
        }
        for(int i=start; i<=end;i++){
            
            vector<TreeNode*> leftTree = buildTree(start, i-1);
            vector<TreeNode*> rightTree = buildTree(i+1, end);
            for(int left=0;left<leftTree.size();left++){
                for(int right=0;right<rightTree.size();right++){
                    TreeNode* node = new TreeNode(i);
                    node->left = leftTree[left];
                    node->right = rightTree[right];
                    tree.push_back(node);
                }
            }
        }
        return tree;
    }
};

Note

To solve the problem, recursion is used for find the left and right child of the tree.


LeetCode Unqiue Binary Search Trees 2
http://zhaoshuaijiang.com/2015/08/13/leetcode_unqiue_binary_search_trees2/
作者
shuaijiang
发布于
2015年8月13日
许可协议