[LeetCode] 122: Unique Binary Search Trees II-优快云博客

本文链接：https://blog.youkuaiyun.com/raojunyang/article/details/77925286

本文介绍了一个算法问题，即如何生成所有可能的唯一二叉搜索树（BST），这些树存储从1到n的值。通过递归方法实现，利用C++代码详细展示了如何构造这些BST，并提供了序列化的例子。

[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:

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);
	}
};