Problem solution · Python

Find All the Lonely Nodes

Find All the Lonely Nodes: a Python solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Depth-first search
Source
Kamyu LeetCode Solutions
Length
54 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 Find All the Lonely Nodes, 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

  • 54 lines of Python from the credited upstream file find-all-the-lonely-nodes.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 Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Full codeFind All the Lonely Nodes · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(n)# Space: O(h) # Definition for a binary tree node.class TreeNode(object):    def __init__(self, val=0, left=None, right=None):        self.val = val        self.left = left        self.right = right  class Solution(object):    def getLonelyNodes(self, root):        """        :type root: TreeNode        :rtype: List[int]        """        result = []        stk = [root]        while stk:            node = stk.pop()            if not node:                continue            if node.left and not node.right:                result.append(node.left.val)            elif node.right and not node.left:                result.append(node.right.val)            stk.append(node.right)            stk.append(node.left)        return result  # Time:  O(n)# Space: O(h)class Solution2(object):    def getLonelyNodes(self, root):        """        :type root: TreeNode        :rtype: List[int]        """        def dfs(node, result):            if not node:                return            if node.left and not node.right:                result.append(node.left.val)            elif node.right and not node.left:                result.append(node.right.val)            dfs(node.left, result)            dfs(node.right, result)         result = []        dfs(root, result)        return result 

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