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95. Unique Binary Search Trees II
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Given an integer n, generate all structurally unique BST's (binary search trees) that store values 1 ... n.
Example:
Input: 3
Output:
[
[1,null,3,2],
[3,2,null,1],
[3,1,null,null,2],
[2,1,3],
[1,null,2,null,3]
]
Explanation:
The above output corresponds to the 5 unique BST's shown below:
1 3 3 2 1
\ / / / \ \
3 2 1 1 3 2
/ / \ \
2 1 2 3
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Difficulty:Medium
Total Accepted:114.1K
Total Submissions:342.3K
Contributor:LeetCode
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Related Topics
Dynamic ProgrammingTree
Similar Questions
Unique Binary Search TreesDifferent Ways to Add Parentheses
Java
right
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/**
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* Definition for a binary tree node.
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* public class TreeNode {
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* int val;
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* TreeNode left;
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* TreeNode right;
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* TreeNode(int x) { val = x; }
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* }
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*/
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class Solution {
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public List<TreeNode> generateTrees(int n ){
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return generateTrees(1,n);
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}
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public List<TreeNode> generateTrees(int m, int n) {
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if (m == n) {
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ArrayList<TreeNode> treeNodes = new ArrayList<>();
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treeNodes.add(new TreeNode(m));
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return treeNodes;
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}
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List<TreeNode> list = new ArrayList<>();
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for (int i = m; i <= n; i++) {
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List<TreeNode> left = null;
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List<TreeNode> right = null;
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if (i == m) {
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right = generateTrees(i + 1, n);
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} else {
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if (i == n) {
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left = generateTrees(m, i - 1);
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} else {
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right = generateTrees(i + 1, n);
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left = generateTrees(m, i - 1);
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}
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}
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if (left != null && right != null) {
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for (TreeNode treeNode : left) {
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for (TreeNode node : right) {
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TreeNode node1 = new TreeNode(i);
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node1.left = treeNode;
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node1.right = node;
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list.add(node1);
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}
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}
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} else {
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if (left != null) {
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for (TreeNode treeNode : left) {
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TreeNode node1 = new TreeNode(i);
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node1.left = treeNode;
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list.add(node1);
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}
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} else {
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for (TreeNode treeNode : right) {
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TreeNode node1 = new TreeNode(i);
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node1.right = treeNode;
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list.add(node1);
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}
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}
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}
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}
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return list;
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}
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}
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