Lintcode-easy-flatten Binary Tree to Linked List

Flatten a binary tree to a fake "linked list" in pre-order traversal.

Here we use the right pointer in TreeNode as the nextpointer in ListNode.

Example

              1                    1          2    / \             2   5    =>    3  / \   \           3   4   6          4                                           5                                               6

Note

Don ' t forget to mark the left child of each node to null. Or you'll get time limit exceeded or Memory Limit exceeded.

Challenge

Do it in-place without any extra memory.

Recursively, note if the left dial hand tree is null directly flatten the right subtree and then returns, otherwise times out

/*** Definition of TreeNode: * public class TreeNode {* public int val; * Public TreeNode left, right; * PU Blic TreeNode (int val) {* This.val = val; * This.left = This.right = null; *} *}*/ Public classSolution {/**     * @paramroot:a TreeNode, the root of the binary tree *@return: Nothing*/     Public voidFlatten (TreeNode root) {//Write your code here        if(Root = =NULL)            return; if(Root.left = =NULL&& Root.right = =NULL)            return; if(Root.left! =NULL) {flatten (root.left); }                if(Root.right! =NULL) {flatten (root.right); } TreeNode Temp=Root.right; if(Root.left! =NULL) {Root.right=Root.left; Root.left=NULL; TreeNode P=Root;  while(P.right! =NULL) P=P.right; P.right=temp; }                                return; }}

Lintcode-easy-flatten Binary Tree to Linked List