LeetCode 95. Unique Binary Search Trees II 动态演示
比如输入为n, 这道题目就是让返回由1,2,... n的组成的所有二叉排序树,每个树都必须包含这n个数
这是二叉树的排列组合的题目。排列组合经常用DFS来解决。
这道题比如输入为3,也就是求start=1, end=3的所有树,简写为t[1,3]。那么就要考虑这些情况:
//t[a,b]={NULL} if a>b
root为1,left为t[1,0], right为t[2,3] left为{NULL}, right为两个树(见下面解释)
root为2,left为t[1,1], right为t[3,3]
root为3,left为t[1,2], right为t[4,3]
类似,要求t[2,3], 就要考虑下面情况
root为2,left为t[2,1], right为t[3,3], 可以看出只有一个
root为3,left为t[2,2], right为t[4,3], 也只有一个
另外对于DFS,为防止重复计算,可以用map来存储计算过的t[x,y]。这也是经常用到的办法。
class Solution { public: vector<TreeNode*> generateTreesDFS(int start, int end, map<pair<int,int>, vector<TreeNode*>>& vecTreeMap) { vector<TreeNode*> subTree; if(vecTreeMap.find({start, end}) != vecTreeMap.end()) return vecTreeMap[{start,end}]; //a(start) //a(end) //ahd(subTree) if (start > end) subTree.push_back(NULL); else { for (int i = start; i <= end; ++i) { vector<TreeNode*> leftSubTree = generateTreesDFS(start, i - 1, vecTreeMap); vector<TreeNode*> rightSubTree = generateTreesDFS(i + 1, end, vecTreeMap); 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); } } } //dsp } vecTreeMap[{start, end}] = subTree; return subTree; } vector<TreeNode*> generateTrees(int n) { if(n==0) return {}; map<pair<int,int>, vector<TreeNode*>> vecTreeMap; //amap(vecTreeMap, pair<int,int>, vector<TreeNode*>) return generateTreesDFS(1,n, vecTreeMap); } };