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/