Unique Binary Search Trees II leetcode java
2014-08-01 03:47
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题目: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 题解: 这道题比1难的就是不是返回个数,而是返回所有结果。 引用code ganker(http://codeganker.blogspot.com/2014/04/unique-binary-search-trees-ii-leetcode.html)的讲解: ”这道题是求解所有可行的二叉查找树,从Unique Binary Search Trees中我们已经知道,可行的二叉查找树的数量是相应的卡特兰数,不是一个多项式时间的数量级,所以我们要求解所有的树,自然是不能多项式时间内完成的了。算法上还是用求解NP问题的方法来求解,也就是N-Queens中 介绍的在循环中调用递归函数求解子问题。思路是每次一次选取一个结点为根,然后递归求解左右子树的所有结果,最后根据左右子树的返回的所有子树,依次选取 然后接上(每个左边的子树跟所有右边的子树匹配,而每个右边的子树也要跟所有的左边子树匹配,总共有左右子树数量的乘积种情况),构造好之后作为当前树的 结果返回。 “ 这道题的解题依据依然是: 当数组为 1,2,3,4,.. i,.. n时,基于以下原则的BST建树具有唯一性: 以i为根节点的树,其左子树由[1, i-1]构成, 其右子树由[i+1, n]构成。 代码如下:1 public ArrayList<TreeNode> generateTrees(int n) {2 return generateTrees(1, n);//从1作为root开始,到n作为root结束3 }45 private ArrayList<TreeNode> generateTrees(int left, int right){6 ArrayList<TreeNode> res = new ArrayList<TreeNode>();7 if (left > right){8 res.add(null);9 return res;}for (int i = left; i <= right; i++){ArrayList<TreeNode> lefts = generateTrees(left, i-1);//以i作为根节点,左子树由[1,i-1]构成ArrayList<TreeNode> rights = generateTrees(i+1, right);//右子树由[i+1, n]构成for (int j = 0; j < lefts.size(); j++){for (int k = 0; k < rights.size(); k++){TreeNode root = new TreeNode(i);root.left = lefts.get(j);root.right = rights.get(k);res.add(root);//存储所有可能行}}}return res;}
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