95. Unique Binary Search Trees II
2018-01-23 20:26
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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.
我首先注意到1..n是任何BST节点1到n的有序遍历。
所以如果我选择第i个节点作为我的根,左子树将包含元素1到(i-1),右子树将包含元素(i + 1)到n。
我使用递归调用来找回所有可能的左树子树和右树子树,并将它们以各种可能的方式与根结合起来。
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
我首先注意到1..n是任何BST节点1到n的有序遍历。
所以如果我选择第i个节点作为我的根,左子树将包含元素1到(i-1),右子树将包含元素(i + 1)到n。
我使用递归调用来找回所有可能的左树子树和右树子树,并将它们以各种可能的方式与根结合起来。
1 class Solution { 2 public List<TreeNode> generateTrees(int n) { 3 if (n<1) return new ArrayList<TreeNode>(); 4 return gen(1,n); 5 } 6 private List<TreeNode> gen(int start,int end){ 7 List<TreeNode> res = new ArrayList<TreeNode>(); 8 if(start>end){ //越界 9 res.add(null); 10 return res; 11 } 12 if(start==end){ //只有一个点 gen(2,2) 13 res.add(new TreeNode(start)); 14 return res; 15 } 16 List<TreeNode> left,right; 17 for(int i = start;i<=end;i++){ 18 left = gen(start,i-1); 19 right = gen(i+1,end); 20 for(TreeNode l :left){ 21 for(TreeNode r:right){ 22 TreeNode root = new TreeNode(i); 23 root.left=l; 24 root.right=r; 25 res.add(root); 26 } 27 } 28 } 29 return res; 30 } 31 }
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