Problem
Given the root of a binary search tree, rearrange the tree in in-order so that the leftmost node in the tree is now the root of the tree, and every node has no left child and only one right child.
Example 1:
Input: root = [5,3,6,2,4,null,8,1,null,null,null,7,9]
Output: [1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]
Example 2:
Input: root = [5,1,7]
Output: [1,null,5,null,7]
Constraints:
The number of nodes in the given tree will be in the range
[1, 100].0 <= Node.val <= 1000
Solution (Java)
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public TreeNode increasingBST(final TreeNode root) {
final List<TreeNode> list = new LinkedList<>();
traverse(root, list);
for (int i = 1; i < list.size(); i++) {
list.get(i - 1).right = list.get(i);
list.get(i).left = null;
}
return list.get(0);
}
private void traverse(final TreeNode root, final List<TreeNode> list) {
if (root != null) {
traverse(root.left, list);
list.add(root);
traverse(root.right, list);
}
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).