I'm busy playing Dungeons & Fighters, so won't do pick up.

Today I have done leetcode's May LeetCoding Challenge with cpp.

May LeetCoding Challenge 14

Description

Flatten Binary Tree to Linked List

Given the root of a binary tree, flatten the tree into a "linked list":

• The "linked list" should use the same TreeNode class where the right child pointer points to the next node in the list and the left child pointer is always null.
• The "linked list" should be in the same order as a pre-order traversal of the binary tree.

Example 1:

Input: root = [1,2,5,3,4,null,6]
Output: [1,null,2,null,3,null,4,null,5,null,6]

Example 2:

Input: root = []
Output: []

Example 3:

Input: root = [0]
Output: [0]

Constraints:

• The number of nodes in the tree is in the range [0, 2000].
• -100 <= Node.val <= 100

Follow up: Can you flatten the tree in-place (with O(1) extra space)?

Solution

class Solution {
  TreeNode* rightMost = nullptr;
public:
  void flatten(TreeNode* root) {
    if(!root) return;
    rightMost = root;
    if(root->left) {
      flatten(root->left);
      rightMost->right = root->right;
      root->right = root->left;
      root->left = nullptr;
    }
    flatten(rightMost->right);
  }
};

class Solution {
public:
  void flatten(TreeNode* root) {
    if(!root) return;
    TreeNode *cur = root;
    TreeNode *prev = nullptr;
    while(cur) {
      if(!cur->left) {
        prev = cur;
        cur = cur->right;
      } else {
        TreeNode *right = cur->left;
        while(right->right != nullptr && right->right != cur) {
          right = right->right;
        }
        if(right->right == cur) {
          right->right = cur->right;
          cur->right = cur->left;
          cur->left = nullptr;
          cur = right->right;
        } else {
          right->right = cur;
          cur = cur->left;
        }
      }
    }
  }
};