Binary Tree Upside Down

LeetCode #156

Description:

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.

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

Idea:

仔细看Code吧，就是root->left变成了新root，左为现在root的right，右为现root。然后把老root的两个子树设为空。

Code:

class Solution {
public:
    TreeNode* upsideDownBinaryTree(TreeNode* root) {
        if(root==NULL) return NULL; // if the tree is empty

        // take care of the leaf
        if(root->left==NULL && root->right==NULL) return root; 

        TreeNode *new_root = upsideDownBinaryTree(root->left);

        root->left->left = root->right;
        root->left->right = root;

        root->left=NULL;
        root->right=NULL;

        return new_root;
    }
};