Unique Binary Search Trees,Unique Binary Search Trees2 生成二叉排序树

Unique Binary Search Trees：求生成二叉排序树的个数。

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

算法分析：类似上阶梯，简单的动态规划问题。当根节点为i时，比i小的节点有i-1个，比i大的节点有n-i个，所以，i为根节点能够生成二叉排序树的个数是

nums[n] += nums[i-1]*nums[n-i]，i从1到n。

public class UniqueBinarySearchTrees
{public int numTrees(int n){if(n <= 0){return 0;}int[] res = new int[n+1];res[0] = 1;res[1] = 1;for(int i = 2; i <= n; i ++){for(int j = 1; j <= i; j ++)//j为根节点{res[i] += res[j-1]*res[i-j];}}return res[n];}
}

Unique Binary Search Trees2：求生成二叉排序树的根节点的集合

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

算法分析：这个不是求个数，而是求生成树根节点。使用递归。

public class UniqueBinarySearchTreesII
{public List<TreeNode> generateTrees(int n) {if(n <= 0){return new ArrayList<TreeNode>();}return helper(1, n);}public List<TreeNode> helper(int m, int n){List<TreeNode> res = new ArrayList<>();if(m > n){res.add(null);return res;}for(int i = m; i <= n; i ++){//i为根节点List<TreeNode> ls = helper(m, i-1);//i节点的左子树List<TreeNode> rs = helper(i+1, n);//i节点的右子树for(TreeNode l : ls){for(TreeNode r : rs){TreeNode curr = new TreeNode(i);curr.left = l;curr.right = r;res.add(curr);}}}return res;}
}

转载于:https://www.cnblogs.com/masterlibin/p/5827259.html

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