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 next pointer in ListNode.

For example,
Given

         1        /        2   5      /\        3   4   6

The flattened tree should look like:

   1         2             3                 4                     5                         6

Analysis:

Put all the tree nodes in the order of pre-order in ArrayList, then traverse the nodes in the ArrayList and adjust their left and right child pointers.

1 /**2 * Definition of TreeNode:3 * public class TreeNode {4 * public int val;5 * Public TreeNode left and right;6 * Public TreeNode (int val) {7 * This.val = val;8 * This.left = This.right = null;9  *     }Ten  * } One  */ A  Public classSolution { -     /** - * @param root:a TreeNode, the root of the binary tree the * @return: Nothing - * Cnblogs.com/beiyeqingteng -      */ -      Public voidFlatten (TreeNode root) { +         //Write your code here -arraylist<treenode> list =NewArraylist<treenode>(); + Preorder (root, list); A          at          for(inti =0; I < list.size (); i++) { -List.Get(i). Left =NULL; -             if(i = = List.size ()-1) { -List.Get(i). Right =NULL; -}Else { -List.Get(i). Right = list.Get(i +1); in             } -         } to     } +      -      Public voidPreorder (TreeNode root, arraylist<treenode>list) { the         if(Root! =NULL) { * List.add (root); $ Preorder (root.left, list);Panax Notoginseng Preorder (root.right, list); -         } the     } +}

View Code

There is a recursive way, the following method refers to another netizen's approach, that this method is very easy to understand, and has a general.

Solution 2: Recursive construction

Suppose that the left and right subtree T (root->left) and T (root->right) of a node have been flatten into the linked list:

1
/    \
2 5
\       \
3 6 <-Righttail
\
4 <-Lefttail

How do I flatten root, t (root->left), T (Root->right) into an entire linked list? Obvious:

temp = Root->right
Root->right = Root->left
Root->left = NULL
Lefttail->right = Temp

We need to use the flatten linked list's trailing element Lefttail, Righttail. So the recursive return value should be the tail of the entire linked list that was generated.

1 classSolution {2      Public voidFlatten (TreeNode root) {3 FLATTENBT (root);4     }5 6 TreeNode flattenbt (TreeNode root) {7         if(Root = =NULL)return NULL;8TreeNode Lefttail =FLATTENBT (root.left);9TreeNode Righttail =FLATTENBT (root.right);Ten if (root.left! = null) { OneTreeNode temp =Root.right; ARoot.right =Root.left; -Root.left =NULL; -Lefttail.right =temp; the         } -         if (righttail! = null) return RIGHTTAIL;18 if (lefttail! = null) return lefttail;19 return ro OT; -     } +}

Several details to be noted
ln 10: Insert the left dial hand tree into the right subtree only if it exists
ln 17-19: When the tail element is returned, special treatment (1) is required to have the right subtree, (2) There is no right subtree but there is a case of Zuozi, (3) The absence of the subtree.

Reference:

Http://bangbingsyb.blogspot.com/2014/11/leetcode-flatten-binary-tree-to-linked.html

Flatten Binary Tree to Linked List