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
难度:medium
题目:给出整数n,生成其所有可能BST.
思路:递归,分别把1到n作为根结点,例如i, 则1到i-1作为左子树,i+1到n作为右子树。
Runtime: 2 ms, faster than 89.21% of Java online submissions for Unique Binary Search Trees II.
Memory Usage: 39.9 MB, less than 100.00% of Java online submissions for Unique Binary Search Trees II.
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<TreeNode> generateTrees(int n) {
if (n <= 0) {
return new ArrayList<TreeNode>();
}
return generateTrees(1, n);
}
private List<TreeNode> generateTrees(int left, int right) {
if (left > right) {
List<TreeNode> treeNodes = new ArrayList<>();
treeNodes.add(null);
return treeNodes;
}
List<TreeNode> result = new ArrayList<>();
for (int i = left; i <= right; i++) {
List<TreeNode> leftList = generateTrees(left, i - 1);
List<TreeNode> rightList = generateTrees(i + 1, right);
for (TreeNode lChild: leftList) {
for (TreeNode rChild: rightList) {
TreeNode root = new TreeNode(i);
root.left = lChild;
root.right = rChild;
result.add(root);
}
}
}
return result;
}
}
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