**Binary Tree Upside Down** **LeetCode Problem | LeetCode Problems For Beginners | LeetCode Problems & Solutions | Improve Problem Solving Skills | LeetCode Problems Java | LeetCode Solutions in C++**

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In this post, you will find the solution for the **Binary Tree Upside Down** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Binary Tree Upside Down– LeetCode Problem

Binary Tree Upside Down– LeetCode Problem

**Problem:**

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

Binary Tree Upside Down– LeetCode Solutions

Binary Tree Upside Down Solution in C++:

class Solution { public: TreeNode* upsideDownBinaryTree(TreeNode* root) { if (!root || !root->left) return root; TreeNode* newRoot = upsideDownBinaryTree(root->left); root->left->left = root->right; // 2's left = 3 (root's right) root->left->right = root; // 2's right = 1 (root) root->left = nullptr; root->right = nullptr; return newRoot; } };

Binary Tree Upside Down Solution in Java:

class Solution { public TreeNode upsideDownBinaryTree(TreeNode root) { if (root == null || root.left == null) return root; TreeNode newRoot = upsideDownBinaryTree(root.left); root.left.left = root.right; // 2's left = 3 (root's right) root.left.right = root; // 2's right = 1 (root) root.left = null; root.right = null; return newRoot; } }

Binary Tree Upside Down Solution in Python:

class Solution: def upsideDownBinaryTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]: if not root or not root.left: return root newRoot = self.upsideDownBinaryTree(root.left) root.left.left = root.right # 2's left = 3 (root's right) root.left.right = root # 2's right = 1 (root) root.left = None root.right = None return newRoot