LeetCode之Unique Binary Search Trees & Unique Binary Search Trees II
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采用Java实现两道题目;
Unique Binary Search Trees题目描述如下:
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题目思路大致如下:对于给定的数n,自1至n分别以每个数作为节点总数计算其Unique Binary Search Trees数目result[i],其中当节点总数为i时的result[i] 计算方法为:
左子树的Unique Binary Search Trees数目 * 右子树的Unique Binary Search Trees,所有情况相加即得result[i],左右子树的节点个数变化范围为(0, i - 1), (1, i - 2)......(i - 1, 0), 代码如下:
<span style="font-family:Arial;font-size:14px;">public class UniqueBinarySearchTrees { public int numTrees(int n) { if (n == 1) return 1; if (n == 2) return 2; int[] result = new int[n + 1]; result[0] = 1; result[1] = 1; result[2] = 2; for (int i = 3; i <= n; i++) { int tmp = 0; for (int k = 0; k < i; k++) { tmp += (result[k] * result[i - k - 1]); } result[i] = tmp; } return result[n]; } }</span>
Unique Binary Search Trees II题目描述如下:
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.
题目相对上题的变化即要返回所有的Unique Binary Search Trees组成的集合,解题思路如下:以自1至n的每个数i作为根节点:比i小的所有节点组成左子树,比i大的所有节点组成右子树,迭代构造得到所有的以i为根节点的Unique Binary Search Tree;然后迭代i,即可构造出所有的Unique Binary Search Tree;代码如下:
public class Solution { public List<TreeNode> generateTrees(int n) { return generateTrees(1, n); } public List<TreeNode> generateTrees(int start, int end){ List<TreeNode> result = new ArrayList<TreeNode>(); if(start > end){ result.add(null); return result; } List<TreeNode> leftTree = new ArrayList<TreeNode>(); List<TreeNode> rightTree = new ArrayList<TreeNode>(); for(int i = start; i <= end; i++){ leftTree = generateTrees(start, i - 1); rightTree = generateTrees(i + 1, end); for(TreeNode left : leftTree) for(TreeNode right : rightTree){ TreeNode root = new TreeNode(i); root.right = right; root.left = left; result.add(root); } } return result; }}
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