LeetCode 题解(232) : 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：

递归，从根节点的左子节点开始处理。

C++版：

class Solution {
public:
    TreeNode* upsideDownBinaryTree(TreeNode* root) {
        if(root == NULL || (root->left == NULL && root->right == NULL))
            return root;
            
        TreeNode* newRoot = upsideDown(root, root->left, root->right);
        root->left = NULL;
        root->right = NULL;
        return newRoot;
    }
    
    TreeNode* upsideDown(TreeNode* root, TreeNode* leftChild, TreeNode* rightChild) {
        if(leftChild->left == NULL) {
            leftChild->left = rightChild;
            leftChild->right = root;
            return leftChild;
        }
        TreeNode* curLeft = leftChild->left;
        TreeNode* curRight = leftChild->right;
        leftChild->left = rightChild;
        leftChild->right = root;
        return upsideDown(leftChild, curLeft, curRight);
    }
};

Java版：

public class Solution {
    public TreeNode upsideDownBinaryTree(TreeNode root) {
        if(root == null || (root.left == null && root.right == null)) {
            return root;
        }
        TreeNode newRoot = upsideDown(root, root.left, root.right);
        root.left = null;
        root.right = null;
        return newRoot;
    }
    
    private TreeNode upsideDown(TreeNode root, TreeNode leftChild, TreeNode rightChild) {
        if(leftChild.left == null) {
            leftChild.left = rightChild;
            leftChild.right = root;
            return leftChild;
        }
        TreeNode curLeft = leftChild.left;
        TreeNode curRight = leftChild.right;
        leftChild.left = rightChild;
        leftChild.right = root;
        return upsideDown(leftChild, curLeft, curRight);
    }
}

Python版：

class Solution(object):
    def upsideDownBinaryTree(self, root):
        """
        :type root: TreeNode
        :rtype: TreeNode
        """
        if root == None or (root.left == None and root.right == None):
            return root
            
        newRoot = self.upsideDown(root, root.left, root.right)
        root.left = None
        root.right = None
        return newRoot
        
    def upsideDown(self, root, leftChild, rightChild):
        if leftChild.left == None:
            leftChild.left = rightChild
            leftChild.right = root
            return leftChild
        curLeft = leftChild.left
        curRight = leftChild.right
        newRoot = leftChild
        newRoot.left = rightChild
        newRoot.right = root
        return self.upsideDown(leftChild, curLeft, curRight)