Flatten Binary Tree to Linked List LeetCode Solution

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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

Flatten Binary Tree to Linked List Solutions

Time: O(n)
Space: O(h)

C++

class Solution {
 public:
  void flatten(TreeNode* root) {
    if (!root)
      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;
  }
};
 

Java

 
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;
  }
}

Python

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

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