[LeetCode] 122: Unique Binary Search Trees II

[Problem]

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

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:

   1  / \ 2   3    /   4    \     5

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

[Solution]

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
// generate BST between [begin, end]
vector<TreeNode *>generateTrees(int begin, int end){
vector<TreeNode *> res;

// invalid
if(begin > end){
res.push_back(NULL);
return res;
}
// only one node
else if(begin == end){
res.push_back(new TreeNode(begin));
return res;
}
else{
// iterate root of the tree
for(int i = begin; i <= end; ++i){
vector<TreeNode *> left = generateTrees(begin, i-1);
vector<TreeNode *> right = generateTrees(i+1, end);

// add result
for(int j = 0; j < left.size(); ++j){
for(int k = 0; k < right.size(); ++k){
TreeNode *root = new TreeNode(i);
root->left = left[j];
root->right = right[k];
res.push_back(root);
}
}
}
return res;
}
}
vector<TreeNode *> generateTrees(int n) {
// Note: The Solution object is instantiated only once and is reused by each test case.
return generateTrees(1, n);
}
};

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