LeetCode 95. Unique Binary Search Trees II
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这道题是一道关于Tree的题目。但是表示把所有情况的放入到一个vector里,我是真的没想好怎么搞。于是找了一个答案,豁然开朗。
把所有左右子树,放入到一个vector里。最后把根节点放入到一个vector,是vector递归的递归。思路很巧妙。
如何处理二叉搜索树,对于子树也满足左子树小于根节点,右子树也大于根节点,这个很简单。只要start小于end就要,不是得就剪枝掉。
代码如下:
/** * 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 *> createTree(int start, int end) { vector<TreeNode *> results; if(start>end) { results.push_back(NULL); return results; } for(int k=start;k<=end;k++) { vector<TreeNode *> left = createTree(start,k-1); vector<TreeNode *> right = createTree(k+1,end); for(int i=0;i<left.size();i++) { for(int j=0;j<right.size();j++) { TreeNode * root = new TreeNode(k); root->left = left[i]; root->right = right[j]; results.push_back(root); } } } return results; } vector<TreeNode*> generateTrees(int n) { vector<TreeNode*>tmp; if(n==0)return tmp; return createTree(1,n); }};
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