The data structure TreeNode is used for binary tree, but it can also used to represent a single linked list (where left is null, and right is the next node in the list). Implement a method to convert a binary search tree (implemented with TreeNode) into a single linked list. The values should be kept in order and the operation should be performed in place (that is, on the original data structure).
Return the head node of the linked list after converting.
Note: This problem is slightly different from the original one in the book.
Example:
Input: [4,2,5,1,3,null,6,0] Output: [0,null,1,null,2,null,3,null,4,null,5,null,6]
Note:
- The number of nodes will not exceed 100000.
See 897. Increasing Order Search Tree.
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def convertBiNode(self, root: TreeNode) -> TreeNode:
if root is None:
return None
left = self.convertBiNode(root.left)
right = self.convertBiNode(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(int x) { val = x; }
* }
*/
class Solution {
public TreeNode convertBiNode(TreeNode root) {
if (root == null) return null;
TreeNode left = convertBiNode(root.left);
TreeNode right = convertBiNode(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;
}
}