1. 程式人生 > >leetcode 95. Unique Binary Search Trees II【求二叉搜尋樹的各種具體情況】

leetcode 95. Unique Binary Search Trees II【求二叉搜尋樹的各種具體情況】

https://leetcode.com/problems/unique-binary-search-trees-ii/

Given an integer n, generate all structurally unique BST's (binary search trees) that store values 1 ... n.

Example:

Input: 3
Output:
[
  [1,null,3,2],
  [3,2,null,1],
  [3,1,null,null,2],
  [2,1,3],
  [1,null,2,null,3]
]
Explanation:
The above output corresponds to the 5 unique BST's shown below: 1 3 3 2 1 \ / / / \ \ 3 2 1 1 3 2 / / \ \ 2 1 2 3

和上一篇類似

用分治的思想

每個點列舉作為根節點,遞迴列舉左子樹,右子樹

注意主函式判斷0直接返回[] 否則返回的是[[]]

TreeNode* root=new TreeNode(i);為什麼只能放在迴圈裡啊……

/**
 * 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*> dfs(int l,int r){
        vector<TreeNode*>ans;
        if(l>r){
            ans.push_back(NULL);
            return ans;
        }
        for(int i=l;i<=r;i++){
            vector<TreeNode*>ansL=dfs(l,i-1);
            vector<TreeNode*>ansR=dfs(i+1,r);
            
            for(int j=0;j<ansL.size();j++){
                for(int k=0;k<ansR.size();k++){
                    TreeNode* root=new TreeNode(i);
                    root->left=ansL[j];
                    root->right=ansR[k];
                    ans.push_back(root);
                }
            }
        }
        return ans;
        
    }
    vector<TreeNode*> generateTrees(int n) {
        vector<TreeNode*> r;
        if(n==0)
            return r;
        return dfs(1,n);
    }
};