156. 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},

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

链接： http://leetcode.com/problems/binary-tree-upside-down/

6/14/2017

1ms, 10%

我以为自己做不出来，结果居然很快做出来了，厚积薄发了嘿嘿。思路是，自底向上，left-root-right变为oldRight-oldLeft-oldRoot，并且上层的parent和right放在新的左子树的最右边节点。

注意，第17，18行需要将原来root的左右指针设为null

问题，第12行被使用过吗？还是说在题目给定条件下不会被用到？

希望自己能准确判断和使用top-down, bottom-up

 1 public class Solution {
 2     public TreeNode upsideDownBinaryTree(TreeNode root) {
 3         if (root == null) {
 4             return null;
 5         }
 6         
 7         if (root.left == null && root.right == null) {
 8             return root;
 9         }
10         
11         TreeNode newLeft = upsideDownBinaryTree(root.left);
12         TreeNode newRight = upsideDownBinaryTree(root.right);
13         setNewUpsideDownTree(newLeft, root, newRight);
14         return newLeft;
15     }
16     private void setNewUpsideDownTree(TreeNode newRoot, TreeNode oldRoot, TreeNode newRight) {
17         oldRoot.left = null;
18         oldRoot.right = null;
19         while (newRoot.right != null) {
20             newRoot = newRoot.right;
21         }
22         newRoot.right = oldRoot;
23         newRoot.left = newRight;
24     }
25 }

别人的做法，包括recursive和iterative

https://discuss.leetcode.com/topic/40924/java-recursive-o-logn-space-and-iterative-solutions-o-1-space-with-explanation-and-figure

recursive

public TreeNode upsideDownBinaryTree(TreeNode root) {
    if(root == null || root.left == null) {
        return root;
    }
    
    TreeNode newRoot = upsideDownBinaryTree(root.left);
    root.left.left = root.right;   // node 2 left children
    root.left.right = root;         // node 2 right children
    root.left = null;
    root.right = null;
    return newRoot;
}

iterative

prev, temp相当于2个来自之前层的指针，比其他linkedlist多了一个prev，next指向下一层将要成为root的节点

 1 public TreeNode upsideDownBinaryTree(TreeNode root) {
 2     TreeNode curr = root;
 3     TreeNode next = null;
 4     TreeNode temp = null;
 5     TreeNode prev = null;
 6     
 7     while(curr != null) {
 8         next = curr.left;
 9         
10         // swapping nodes now, need temp to keep the previous right child
11         curr.left = temp;
12         temp = curr.right;
13         curr.right = prev;
14         
15         prev = curr;
16         curr = next;
17     }
18     return prev;
19 }