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LeetCode Solutions

114. Flatten Binary Tree to Linked List

Time: $O(n)$

Space: $O(h)$

			


class Solution {
 public:
  void flatten(TreeNode* root) {
    if (root == nullptr)
      return;

    flatten(root->left);
    flatten(root->right);

    TreeNode* const left = root->left;    // Flattened left
    TreeNode* const right = root->right;  // Flattened right

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

    // Connect the original right subtree
    // To the end of new right subtree
    TreeNode* rightmost = root;
    while (rightmost->right)
      rightmost = rightmost->right;
    rightmost->right = right;
  }
};




			


class Solution {
  public void flatten(TreeNode root) {
    if (root == null)
      return;

    flatten(root.left);
    flatten(root.right);

    TreeNode left = root.left;   // Flattened left
    TreeNode right = root.right; // Flattened right

    root.left = null;
    root.right = left;

    // Connect the original right subtree
    // To the end of new right subtree
    TreeNode rightmost = root;
    while (rightmost.right != null)
      rightmost = rightmost.right;
    rightmost.right = right;
  }
}




			


class Solution:
  def flatten(self, root: Optional[TreeNode]) -> None:
    if not root:
      return

    self.flatten(root.left)
    self.flatten(root.right)

    left = root.left  # Flattened left
    right = root.right  # Flattened right

    root.left = None
    root.right = left

    # Connect the original right subtree
    # To the end of new right subtree
    rightmost = root
    while rightmost.right:
      rightmost = rightmost.right
    rightmost.right = right