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114. Flatten Binary Tree to Linked List 👍

Approach 1: Recursive

  • Time: $O(n)$
  • Space: $O(h)$
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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 the new right subtree.
    TreeNode* rightmost = root;
    while (rightmost->right != nullptr)
      rightmost = rightmost->right;
    rightmost->right = right;
  }
};

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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 the new right subtree.
    TreeNode rightmost = root;
    while (rightmost.right != null)
      rightmost = rightmost.right;
    rightmost.right = right;
  }
}

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class Solution:
  def flatten(self, root: TreeNode | None) -> 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 the new right subtree.
    rightmost = root
    while rightmost.right:
      rightmost = rightmost.right
    rightmost.right = right

Approach 2: Iterative (stack)

  • Time: $O(n)$
  • Space: $O(h)$
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class Solution {
 public:
  void flatten(TreeNode* root) {
    if (root == nullptr)
      return;

    stack<TreeNode*> stack{{root}};

    while (!stack.empty()) {
      root = stack.top(), stack.pop();
      if (root->right)
        stack.push(root->right);
      if (root->left)
        stack.push(root->left);
      if (!stack.empty())
        root->right = stack.top();
      root->left = nullptr;
    }
  }
};

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

    Deque<TreeNode> stack = new ArrayDeque<>();
    stack.push(root);

    while (!stack.isEmpty()) {
      root = stack.pop();
      if (root.right != null)
        stack.push(root.right);
      if (root.left != null)
        stack.push(root.left);
      if (!stack.isEmpty())
        root.right = stack.peek();
      root.left = null;
    }
  }
}

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

    stack = [root]

    while stack:
      root = stack.pop()
      if root.right:
        stack.append(root.right)
      if root.left:
        stack.append(root.left)
      if stack:
        root.right = stack[-1]
      root.left = None

Approach 3: Morris-like

  • Time: $O(n)$
  • Space: $O(1)$
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class Solution {
 public:
  void flatten(TreeNode* root) {
    if (root == nullptr)
      return;

    while (root != nullptr) {
      if (root->left != nullptr) {
        // Find the rightmost root.
        TreeNode* rightmost = root->left;
        while (rightmost->right != nullptr)
          rightmost = rightmost->right;
        // Rewire the connections.
        rightmost->right = root->right;
        root->right = root->left;
        root->left = nullptr;
      }
      // Move on to the right side of the tree.
      root = root->right;
    }
  }
};

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

    while (root != null) {
      if (root.left != null) {
        // Find the rightmost root.
        TreeNode rightmost = root.left;
        while (rightmost.right != null)
          rightmost = rightmost.right;
        // Rewire the connections.
        rightmost.right = root.right;
        root.right = root.left;
        root.left = null;
      }
      // Move on to the right side of the tree.
      root = root.right;
    }
  }
}

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

    while root:
      if root.left:
        # Find the rightmost root
        rightmost = root.left
        while rightmost.right:
          rightmost = rightmost.right
        # Rewire the connections
        rightmost.right = root.right
        root.right = root.left
        root.left = None
      # Move on to the right side of the tree
      root = root.right