156. Binary Tree Upside Down

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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:

  • The original left child becomes the new root.
  • The original root becomes the new right child.
  • 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:

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

Example 2:

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Input: root = []
Output: []

Example 3:

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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.

Hints/Notes

  • 2025/02/11 Q3
  • binary tree
  • No solution from 0x3F or Leetcode

Solution

Language: C++

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/**
 * 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) {
            return root;
        }
        if (!root->left) {
            return root;
        }
        return helper(root->left, root, root->right);
    }

    TreeNode* helper(TreeNode* left, TreeNode* parent, TreeNode* right) {
        TreeNode* newHead = left;
        if (left->left) {
            newHead = helper(left->left, left, left->right);
        }
        parent->left = nullptr;
        parent->right = nullptr;
        left->left = right;
        left->right = parent;
        return newHead;
    }
};