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156. Binary Tree Upside Down

Description

Given the root of a binary tree, turn the tree upside down and return the new root.

You can turn a binary tree upside down with the following steps:

1. The original left child becomes the new root.
2. The original root becomes the new right child.
3. The original right child becomes the new left child.

The mentioned steps are done level by level. It is guaranteed that every right node has a sibling (a left node with the same parent) and has no children.

 

Example 1:

Input: root = [1,2,3,4,5]
Output: [4,5,2,null,null,3,1]

Example 2:

Input: root = []
Output: []

Example 3:

Input: root = [1]
Output: [1]

 

Constraints:

• The number of nodes in the tree will be in the range [0, 10].
• 1 <= Node.val <= 10
• Every right node in the tree has a sibling (a left node that shares the same parent).
• Every right node in the tree has no children.

Solutions

• Java
• C++
• Python
• Go

• /**
   * Definition for a binary tree node.
   * public class TreeNode {
   *     int val;
   *     TreeNode left;
   *     TreeNode right;
   *     TreeNode() {}
   *     TreeNode(int val) { this.val = val; }
   *     TreeNode(int val, TreeNode left, TreeNode right) {
   *         this.val = val;
   *         this.left = left;
   *         this.right = right;
   *     }
   * }
   */
  class Solution {
      public TreeNode upsideDownBinaryTree(TreeNode root) {
          if (root == null || root.left == null) {
              return root;
          }
          TreeNode newRoot = upsideDownBinaryTree(root.left);
          root.left.right = root;
          root.left.left = root.right;
          root.left = null;
          root.right = null;
          return newRoot;
      }
  }
  

• /**
   * Definition for a binary tree node.
   * struct TreeNode {
   *     int val;
   *     TreeNode *left;
   *     TreeNode *right;
   *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
   *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
   *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
   * };
   */
  class Solution {
  public:
      TreeNode* upsideDownBinaryTree(TreeNode* root) {
          if (!root || !root->left) return root;
          TreeNode* newRoot = upsideDownBinaryTree(root->left);
          root->left->right = root;
          root->left->left = root->right;
          root->left = nullptr;
          root->right = nullptr;
          return newRoot;
      }
  };
  

• # Definition for a binary tree node.
  # class TreeNode:
  #     def __init__(self, val=0, left=None, right=None):
  #         self.val = val
  #         self.left = left
  #         self.right = right
  class Solution:
      def upsideDownBinaryTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
          if root is None or root.left is None:
              return root
          new_root = self.upsideDownBinaryTree(root.left)
          root.left.right = root
          root.left.left = root.right
          root.left = None
          root.right = None
          return new_root
  
  

• /**
   * Definition for a binary tree node.
   * type TreeNode struct {
   *     Val int
   *     Left *TreeNode
   *     Right *TreeNode
   * }
   */
  func upsideDownBinaryTree(root *TreeNode) *TreeNode {
  	if root == nil || root.Left == nil {
  		return root
  	}
  	newRoot := upsideDownBinaryTree(root.Left)
  	root.Left.Right = root
  	root.Left.Left = root.Right
  	root.Left = nil
  	root.Right = nil
  	return newRoot
  }
  

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