Binary Tree Upside Down

Given a binary tree where all the right nodes are either leaf nodes with a sibling (a left node that shares the same parent node) or empty, flip it upside down and turn it into a tree where the original right nodes turned into left leaf nodes. Return the new root.

For example:
Given a binary tree {1,2,3,4,5},

    1

   / \

  2   3

 / \

4   5

 

return the root of the binary tree [4,5,2,#,#,3,1].

   4

  / \

 5   2

    / \

   3   1  

 

将left node 入stack， 然后将每个原来的left node的 left = parent.right, right = parent. (从下往上做)

 1 /**

 2  * Definition for binary tree

 3  * public class TreeNode {

 4  *     int val;

 5  *     TreeNode left;

 6  *     TreeNode right;

 7  *     TreeNode(int x) { val = x; }

 8  * }

 9  */

10 public class Solution {

11     public TreeNode UpsideDownBinaryTree(TreeNode root) {

12         if(root == null) return root;

13         LinkedList<TreeNode> ll = new LinkedList<TreeNode> ();

14         ll.push(root);

15         while(root.left != null){

16             ll.push(root.left);

17             root = root.left;

18         }

19         while(!ll.isEmpty()){

20             TreeNode tmp = ll.pop();

21             tmp.right = ll.peek();

22             tmp.left = ll.peek() == null ? null : ll.peek().right;

23         }

24         return root;

25     }

26 }

 

 从上往下做：

 1 public class Solution {

 2     public TreeNode UpsideDownBinaryTree(TreeNode root) {

 3         if(root == null) return root;

 4         TreeNode cur = root, last = null, lastRight = null;

 5         while(cur != null){

 6             TreeNode tmp = cur.left;

 7             cur.left = lastRight;

 8             lastRight = cur.right;

 9             cur.right = last;

10             last = cur;

11             cur = tmp;

12         }

13         return last;

14     }

15 }