LeetCode——Unique Binary Search Trees II-优快云博客

本文链接：https://blog.youkuaiyun.com/pickless/article/details/11872543

本文介绍了一种使用深度优先搜索(DFS)策略来生成所有可能的独特二叉搜索树的方法。当输入为n时，该算法能找出所有结构上不同的二叉搜索树，这些树将存储从1到n的值。

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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.

» Solve this problem

DFS搜索出所有解。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
private:
    vector<TreeNode *> generate(int l, int r) {
        vector<TreeNode *> ans;
        if (l > r) {
            ans.push_back(NULL);        
        }    
        else {
            for (int i = l; i <= r; i++) {
                vector<TreeNode *> left = generate(l, i - 1);
                vector<TreeNode *> right = generate(i + 1, r);
                for (int idxL = 0; idxL < left.size(); idxL++) {
                    for (int idxR = 0; idxR < right.size(); idxR++) {
                        TreeNode *root = new TreeNode(i);
                        root->left = left[idxL];
                        root->right = right[idxR];
                        ans.push_back(root);
                    }
                }
            }
        }
        return ans;
    }
    
public:
    vector<TreeNode *> generateTrees(int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<TreeNode *> ans = generate(1, n);
        return ans;
    }
};