156. Binary Tree Upside Down

 
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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

 

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.


OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1
  / \
 2   3
    /
   4
    \
     5
The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".
 
给定一棵树，这棵树的右节点有两种选择：1、空节点 2、叶子结点（左节点一定存在）
 
然后旋转该树（结构对称）且左节点变换为：1、空节点 2、叶子结点（右节点一定存在）
 
那么就可以用递归和非递归两种形式实现。
 
我选择了非递归的实现方法。
 
自顶向下：给定[1,2,3]，变换为[2,3,1]依次从上向下变换即可。
 
/**
 * 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 && root.right == null)){
            return root;
        }
        TreeNode node = root;
        TreeNode nodeLeft = root.left;
        TreeNode nodeRight = root.right;
        while (nodeLeft != null){
            TreeNode nodeChange = node;
            TreeNode nodeLeftChange = nodeLeft;
            TreeNode nodeRightChange = nodeRight;
            node = nodeLeft;
            nodeLeft = node.left;
            nodeRight = node.right;
            nodeLeftChange.left = nodeRightChange;
            nodeLeftChange.right = nodeChange;
        }
        root.left = null;
        root.right = null;
        return node;
    }
}