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Problem solution · Python

Boundary of Binary Tree

Boundary of Binary Tree: a Python solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Depth-first search
Source
walkccc LeetCode Solutions
Length
32 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Depth-first search

For Boundary of Binary Tree, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 32 lines of Python from the credited upstream file 545.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeBoundary of Binary Tree · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def boundaryOfBinaryTree(self, root: TreeNode | None) -> list[int]:    if not root:      return []     ans = [root.val]     def dfs(root: TreeNode | None, lb: bool, rb: bool):      """      1. root.left is left boundary if root is left boundary.         root.right if left boundary if root.left is None.      2. Same applys for right boundary.      3. If root is left boundary, add it before 2 children - preorder.         If root is right boundary, add it after 2 children - postorder.      4. A leaf that is neighter left/right boundary belongs to the bottom.      """      if not root:        return      if lb:        ans.append(root.val)      if not lb and not rb and not root.left and not root.right:        ans.append(root.val)       dfs(root.left, lb, rb and not root.right)      dfs(root.right, lb and not root.left, rb)      if rb:        ans.append(root.val)     dfs(root.left, True, False)    dfs(root.right, False, True)    return ans

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