Leetcode: Unique Binary Search Trees II_given n generate all structurally unique bst鈥檚 (bi-优快云博客

本文链接：https://blog.youkuaiyun.com/doc_sgl/article/details/12348041

本文详细介绍了如何生成所有结构独特的n个元素的二叉搜索树，包括二叉搜索树的定义、序列化表示、以及生成算法实现。重点展示了通过递归方法构建所有可能的树结构，确保每棵树都是唯一的。

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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

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

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}".

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<TreeNode *> generate(int left, int right)
    {
		vector<TreeNode *> res;
		if(left>right)return res;
		if(left == right)
		{
			res.push_back(new TreeNode(left));
			return res;
		}
		for(int r = left; r <= right; r++)
		{
			vector<TreeNode*> lchild;
			vector<TreeNode*> rchild;
			if(r==left)
			{
				rchild = generate(left+1,right);
				for(int i =0; i< rchild.size(); i++)
				{
					TreeNode* root = new TreeNode(r);
					root->right = rchild[i];
					res.push_back(root);
				}
			}
			else if(r == right)
			{
				lchild = generate(left, right-1);
				for(int i = 0; i < lchild.size(); i++)
				{
					TreeNode* root = new TreeNode(r);
					root->left = lchild[i];
					res.push_back(root);
				}
			}
			else
			{
				lchild = generate(left,r-1);
				rchild = generate(r+1,right);
				for(int i = 0; i < lchild.size(); i++)
				{
					for(int j = 0; j <rchild.size(); j++)
					{
						TreeNode* root = new TreeNode(r);
						root->left = lchild[i];
						root->right = rchild[j];
						res.push_back(root);
					}
				}
			}
		}
		return res;
	}
	vector<TreeNode *> generateTrees(int n) {
        // Note: The Solution object is instantiated only once.
        if(n<1){
			vector<TreeNode*> res;
			TreeNode * tmp =NULL;
			res.push_back(tmp);
			return res;
		}
        return generate(1,n);
    }
};