LeetCode Unique Binary Search Trees II

最新推荐文章于 2020-12-09 16:51:01 发布

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本文介绍了一种使用深度优先搜索（DFS）的方法来生成所有可能的不同结构的二叉搜索树（BST），这些BST可以存储从1到n的整数值。以n=3为例，展示了如何得到5种不同的BST结构。

题目：

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 binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
//dfs
class Solution {
public:
	vector<TreeNode *> generateTrees(int n) {
		vector<TreeNode *> res = dfs(1, n);
		return res;
	}
private:
	vector<TreeNode *> dfs(int begin, int end) {
		vector<TreeNode *> ans;
		if (begin > end) {
			ans.push_back(NULL);
			return ans;
		}
		for (int i = begin; i <= end; i++) {
			vector<TreeNode *> leftTree = dfs(begin, i - 1);
			vector<TreeNode *> rightTree = dfs(i + 1, end);
			for (int m = 0; m < leftTree.size(); m++) {
				for (int n = 0; n < rightTree.size(); n++) {
					TreeNode *root = new TreeNode(i);
					root->left = leftTree[m];
					root->right = rightTree[n];
					ans.push_back(root);
				}
			}
		}
		return ans;
	}
};