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
See 17.12. BiNode.
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def increasingBST(self, root: TreeNode) -> TreeNode:
if root is None:
return None
left = self.increasingBST(root.left)
right = self.increasingBST(root.right)
if left is None:
root.right = right
return root
res = left
while left and left.right:
left = left.right
left.right = root
root.right = right
root.left = None
return res/**
* 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(TreeNode root) {
if (root == null) return null;
TreeNode left = increasingBST(root.left);
TreeNode right = increasingBST(root.right);
if (left == null) {
root.right = right;
return root;
}
TreeNode res = left;
while (left != null && left.right != null) left = left.right;
left.right = root;
root.right = right;
root.left = null;
return res;
}
}