Binary Tree Upside Down

网上抄的 http://blog.csdn.net/whuwangyi/article/details/43186045
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

一看就懵了，要弄what咧，貌似是，orig.->new: right->left, left ->parent, parent->right, node->parent

直接copy 人家的code

地址
http://blog.csdn.net/whuwangyi/article/details/43186045
“第二个思路是直接用迭代代替递归，做起来也不麻烦，并且效率会更高，因为省去了递归所用的栈空间。

public TreeNode UpsideDownBinaryTree(TreeNode root) {
TreeNode node = root, parent = null, right = null;
while (node != null) {
TreeNode left = node.left;
node.left = right;
right = node.right;
node.right = parent;
parent = node;
node = left;
}
return parent;
}

”