LeetCode Unique Binary Search Trees II

最新推荐文章于 2020-12-09 16:51:01 发布

maqingli87

最新推荐文章于 2020-12-09 16:51:01 发布

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分类专栏： leetcode 文章标签： search serialization traversal tree null struct

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Unique Binary Search Trees II

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}".

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) {
        vector<TreeNode*> ret;
    	dfs(ret, 1, n);
		return ret;
    }

	void dfs(vector<TreeNode*>& ret, int left, int right)
	{
		if(left > right)
			ret.push_back(NULL);
		else
		{
			for(int i = left; i <= right; ++i)
			{
				vector<TreeNode*> lefts;
				dfs(lefts, left, i-1);
				vector<TreeNode*> rights;
				dfs(rights, i+1, right);
				for(int x = 0; x < lefts.size(); ++x)
					for(int y = 0; y < rights.size(); ++y)
					{
						TreeNode* root = new TreeNode(i);
						root->left = lefts[x];
						root->right = rights[y];
						ret.push_back(root);
					}
			}
		}
	}
};