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;
}
}
 
 

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