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
/** * 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) { List<TreeNode> res = new ArrayList<>(); if(n==0){ res.add(null); } for (int i = 1; i <= n; i++) { List<TreeNode> left = generateTrees(1, i - 1); List<TreeNode> right = generateTrees(i + 1, n); for (int k = 0; k < left.size(); k++) { for (int t = 0; t < right.size(); t++) { TreeNode root = new TreeNode(i); root.left = left.get(k); root.right = right.get(t); res.add(root); } } } return res; } private List<TreeNode> generateTrees(int m, int n) { List<TreeNode> res = new ArrayList<>(); if (m > n) { res.add(null); return res; } if (m == n) { TreeNode tn = new TreeNode(m); res.add(tn); return res; } for (int i = m; i <= n; i++) { List<TreeNode> left = generateTrees(m, i - 1); List<TreeNode> right = generateTrees(i + 1, n); for (int k = 0; k < left.size(); k++) { for (int t = 0; t < right.size(); t++) { TreeNode root = new TreeNode(i); root.left = left.get(k); root.right = right.get(t); res.add(root); } } } return res; } }
[LeetCode]Unique Binary Search Trees II
原文地址:http://blog.csdn.net/guorudi/article/details/41390865