Given 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 1 right child.
Example 1: Input: [5,3,6,2,4,null,8,1,null,null,null,7,9]
5 / \ 3 6 / \ \ 2 4 8 / / \ 1 7 9
Output: [1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]
1 \ 2 \ 3 \ 4 \ 5 \ 6 \ 7 \ 8 \ 9
Constraints:
The number of nodes in the given tree will be between 1 and 100. Each node will have a unique integer value from 0 to 1000.
/**
* Definition for a binary tree node.
* public class TreeNode {
* public int val;
* public TreeNode left;
* public TreeNode right;
* public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
public class Solution {
TreeNode prev=null;
TreeNode head=null;
public TreeNode IncreasingBST(TreeNode root) {
if(root==null){
return null;
}
IncreasingBST(root.left);
if(prev != null){
prev.left = null;
prev.right = root;
}
if(head==null){
head = root;
}
prev = root;
IncreasingBST(root.right);
return head;
}
}Time Complexity: O(n)
Space Complexity: O(height)
height is the height of tree.
