LeetCode Binary Tree Upside Down

原题链接在这里：https://leetcode.com/problems/binary-tree-upside-down/

题目：

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  

题解：

Recursion 方法是自底向上.

Time Complexity: O(n).

Space: O(n). tree height.

AC Java:

 /**
  * Definition for a binary tree node.
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 public class Solution {
     public TreeNode upsideDownBinaryTree(TreeNode root) {
         if(root == null || root.left  == null){
             return root;
         }
         TreeNode newRoot = upsideDownBinaryTree(root.left);
 
         root.left.left = root.right;
         root.left.right = root;
 
         root.left = null;
         root.right = null;
         return newRoot;
     }
 }

Iterative 是从上到下.

Time Complexity: O(n). Space: O(1).

AC Java:

 /**
  * Definition for a binary tree node.
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 public class Solution {
     public TreeNode upsideDownBinaryTree(TreeNode root) {
         if(root == null || root.left  == null){
             return root;
         }
         TreeNode cur = root;
         TreeNode next = null;
         TreeNode pre = null;
         TreeNode temp = null;
         while(cur != null){
             next = cur.left;
             cur.left = temp;
             temp = cur.right;
             cur.right = pre;
             pre = cur;
             cur = next;
         }
         return pre;
     }
 }

类似Reverse Linked List.