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