【leetcode】Unique Binary Search Trees II

最新推荐文章于 2020-12-09 16:51:01 发布

Cabinathor

最新推荐文章于 2020-12-09 16:51:01 发布

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分类专栏： Leetcode

本文链接：https://blog.youkuaiyun.com/lmxmimihuhu/article/details/46611467

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本文介绍了一种算法，用于生成所有可能的独特二叉搜索树（BST），这些树可以存储从1到给定整数n的值。通过递归地构建左子树和右子树，并将它们连接到根节点上，该算法确保了树结构的独特性。

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From： https://leetcode.com/problems/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

/**
 * 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) {
       return createTree(1,n);
    }
    
    vector<TreeNode *> createTree(int start, int end) {
        vector<TreeNode *> results;
        if(start > end) {
			results.push_back(NULL);
			return results;
		}
        
        for(int k=start; k<=end; k++) {
            vector<TreeNode *> left = createTree(start, k-1);
            vector<TreeNode *> right = createTree(k+1, end);
            for(int i=0; i<left.size(); i++) {
                for(int j=0; j<right.size(); j++) {
                    TreeNode * root = new TreeNode(k);
                    root->left = left[i];
                    root->right = right[j];
                    results.push_back(root);
                }
            }
        }
        return results;
    }
};

	public List<TreeNode> generateTrees(int n) {
		return construct(1, n);
	}

	private List<TreeNode> construct(int start, int end) {
		List<TreeNode> ans = new ArrayList<TreeNode>();
		if (start > end) {
			ans.add(null);
			return ans;
		}
		for (int k = start; k <= end; ++k) {
			List<TreeNode> left = construct(start, k - 1);
			List<TreeNode> right = construct(k + 1, end);
			for (int i = 0; i < left.size(); ++i) {
				for (int j = 0; j < right.size(); ++j) {
					TreeNode root = new TreeNode(k);
					root.left = left.get(i);
					root.right = right.get(j);
					ans.add(root);
				}
			}
		}
		return ans;
    }