Posts Tagged 【Catalan && dfs】Unique Binary Search Trees I && II

Unique Binary Search Trees

Total Accepted: 47048 Total
Submissions: 129376My Submissions

Question
Solution

Given n, how many structurally unique BST's (binary search trees) that store values 1...n?
For example,

Given n = 3, there are a total of 5 unique BST's.

1         3     3      2      1
\       /     /      / \      \
3     2     1      1   3      2
/     /       \                 \
2     1         2                 3

/*
参考http://blog.csdn.net/csu54zzg/article/details/45148007
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。
i=1,2,3..n，左右两边各有几个
f(0) = 1, f(1) = 1
f(n) = f(0)*f(n-1)+f(1)*f(n-2)+...f(n-1)*f(0)
*/
public class Solution {
public int numTrees(int n) {
if(n<2) return 1;
int[] f = new int[n+1];
f[0] = f[1] = 1;
for(int i = 2;i<n+1;i++) {
f[i] = 0;
for(int j = 0;j<i;j++) {
f[i] += f[j]*f[i-1-j];
}
}
return f
;
}
}

Unique Binary Search Trees II

Total Accepted: 29265 Total
Submissions: 105724My Submissions

Question
Solution

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

confused what

"{1,#,2,3}"

means? >
read more on how binary tree is serialized on OJ.

/*
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。
每一子树返回可能的各种组合根节点
递归, 对于[start, end]范围内的每个节点, 产生所有可能的左、右子树,
再产生(#左子树 x #右子树)棵树, 返回所有的root nodes。
gen函数返回一个list of root nodes,
每个root node所表示的树是由[start, end]这个范围内的数构成的BST。
*/
/**
* Definition for binary tree
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; left = null; right = null; }
* }
*/
public class Solution {
public List<TreeNode> generateTrees(int n) {
return generateTrees(1,n);
}
private List<TreeNode> generateTrees(int start, int end) {
List<TreeNode> list = new ArrayList<TreeNode>();
if(start > end) {
list.add(null);
return list;
}
for(int i = start;i<=end;i++) {
List<TreeNode> treeLeft = new ArrayList<TreeNode>();
List<TreeNode> treeRight = new ArrayList<TreeNode>();
treeLeft = generateTrees(start, i-1);
treeRight = generateTrees(i+1, end);
for(int j = 0;j<treeLeft.size();j++)
for(int k = 0;k<treeRight.size();k++) {
TreeNode root = new TreeNode(i);
root.left = treeLeft.get(j);
root.right = treeRight.get(k);
list.add(root);
}
}
return list;
}
}

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