[LeetCode] Unique Binary Search Trees II 獨一無二的二叉搜尋樹之二
阿新 • • 發佈:2018-12-27
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:
1 / \ 2 3 / 4 \ 5The above binary tree is serialized as
"{1,2,3,#,#,4,#,#,5}"
這道題是之前的 Unique Binary Search Trees 獨一無二的二叉搜尋樹的延伸,之前那個只要求算出所有不同的二叉搜尋樹的個數,這道題讓把那些二叉樹都建立出來。這種建樹問題一般來說都是用遞迴來解,這道題也不例外,劃分左右子樹,遞迴構造。至於遞迴函式中為啥都用的是指標,是參考了網友水中的魚的部落格,若不用指標,全部例項化的話會存在大量的物件拷貝,要呼叫拷貝建構函式,具體我也不太懂,反正感覺挺有道理的,不明覺厲啊-.-!!!
class Solution { public: vector<TreeNode *> generateTrees(intn) { if (n == 0) return {}; return *generateTreesDFS(1, n); } vector<TreeNode*> *generateTreesDFS(int start, int end) { vector<TreeNode*> *subTree = new vector<TreeNode*>(); if (start > end) subTree->push_back(NULL); else { for (int i = start; i <= end; ++i) { vector<TreeNode*> *leftSubTree = generateTreesDFS(start, i - 1); vector<TreeNode*> *rightSubTree = generateTreesDFS(i + 1, end); for (int j = 0; j < leftSubTree->size(); ++j) { for (int k = 0; k < rightSubTree->size(); ++k) { TreeNode *node = new TreeNode(i); node->left = (*leftSubTree)[j]; node->right = (*rightSubTree)[k]; subTree->push_back(node); } } } } return subTree; } };