leetcode95~Unique Binary Search Trees II
2017-02-22 21:56
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Given an integer n, generate all structurally unique BST’s (binary search trees) that store values 1…n.
1 3 3 2 1
\ / / / \ \
3 2 1 1 3 2
/ / \ \
2 1 2 3
但是此题有个坑,网上流传的很多答案都是无法在leetcode中accept的,原因在于输入为0时,返回的结果是[[]],而正确结果是[],这点要格外注意!
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
由上题可知,可行的BST数是相应的卡特兰数,此题是求解所有的树,思路是采用递归。 每次选取一个节点作为根节点,然后递归求解左右子树的结果,最后根据左右子树返回的所有子树,依次选取然后链接即可。
但是此题有个坑,网上流传的很多答案都是无法在leetcode中accept的,原因在于输入为0时,返回的结果是[[]],而正确结果是[],这点要格外注意!
public class UniqueBST95 { public List<TreeNode> generateTrees(int n) { List<TreeNode> res = new ArrayList<TreeNode>(); if(n==0) return res; return helper(1,n); //从节点1开始到节点n,分别作为root节点 } private List<TreeNode> helper(int start, int end) { List<TreeNode> res = new ArrayList<TreeNode>(); if(start>end) { res.add(null); return res; } //以i作为根节点 for(int i=start;i<=end;i++) { List<TreeNode> left = helper(start,i-1); //左子树由start...i-1构成 List<TreeNode> right = helper(i+1,end); //右子树由i+1....end构成 //null也是可以遍历到的 for(TreeNode l : left) { for(TreeNode r : right) { TreeNode root = new TreeNode(i); root.left = l; root.right = r; res.add(root); } } } return res; } }
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