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【LeetCode】Unique Binary Search Trees II
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
由于1~n是升序列,因此建起来的树天然就是BST。
递归思想,依次选择根节点,对左右子序列再分别建树。
由于左右子序列建树的结果也可能不止一种,需要考虑所有搭配情况。
/** * Definition for binary tree * 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) { vector<TreeNode *> res; if(n == 0) { res.push_back(NULL); return res; } else { return Helper(1, n); } } vector<TreeNode*> Helper(int begin, int end) { vector<TreeNode*> res; if(begin > end) { res.push_back(NULL); return res; } else if(begin == end) { TreeNode* root = new TreeNode(begin); res.push_back(root); return res; } else { for(int i = begin; i <= end; i ++) { //tag vector<TreeNode*> left = Helper(begin, i-1); vector<TreeNode*> right = Helper(i+1, end); for(int l = 0; l < left.size(); l ++) {//all BST in left for(int r = 0; r < right.size(); r ++) {//all BST in right //new root here! not the tag position. TreeNode* root = new TreeNode(i); root->left = left[l]; root->right = right[r]; res.push_back(root); } } } return res; } }};
【LeetCode】Unique Binary Search Trees II
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