个人记录-LeetCode 95. Unique Binary Search Trees II
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问题:
Given an integer 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
代码示例:
这个问题比较容易想到的方法是递归。
任取一个数作为根节点;
然后,得到该节点左侧数组成的树集合,以及该节点右侧数组成的树集合;
利用左侧树、根节点及右侧树共同组成最终需要的树。
每个数均重复上述操作,就可以得到最终的树集合。
代码如下:
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */public class Solution { public List<TreeNode> generateTrees(int n) { //n不满足条件,返回空集合 if (n < 1) { return new ArrayList<>(); } //递归获取结果 return innerHelper(1, n); } private List<TreeNode> innerHelper(int start, int end) { List<TreeNode> rst = new ArrayList<>(); if(start > end) { rst.add(null); return rst; } if(start == end){ rst.add(new TreeNode(start)); return rst; } List<TreeNode> left, right; //i作为根节点 for(int i = start; i <= end; i++) { //得到i左侧的数组成的集合 left = innerHelper(start, i-1); //得到i右侧的数组成的集合 right = innerHelper(i+1, end); //左侧和右侧均可能仅包含null for(TreeNode leftNode : left) { for(TreeNode rightNode : right) { //根节点与其左右侧集合组成完整的树 TreeNode root = new TreeNode(i); root.left = leftNode; root.right = rightNode; //保存到结果中 rst.add(root); } } } return rst; }}
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