LeetCode 1469. Find All The Lonely Nodes

Description

https://leetcode.com/problems/find-all-the-lonely-nodes/

In a binary tree, a lonely node is a node that is the only child of its parent node. The root of the tree is not lonely because it does not have a parent node.

Given the root of a binary tree, return an array containing the values of all lonely nodes in the tree. Return the list in any order.

Example 1:

Input: root = [1,2,3,null,4]
Output: [4]
Explanation: Light blue node is the only lonely node.
Node 1 is the root and is not lonely.
Nodes 2 and 3 have the same parent and are not lonely.

Example 2:

Input: root = [7,1,4,6,null,5,3,null,null,null,null,null,2]
Output: [6,2]
Explanation: Light blue nodes are lonely nodes.
Please remember that order doesn't matter, [2,6] is also an acceptable answer.

Example 3:


Input: root = [11,99,88,77,null,null,66,55,null,null,44,33,null,null,22]
Output: [77,55,33,66,44,22]
Explanation: Nodes 99 and 88 share the same parent. Node 11 is the root.
All other nodes are lonely.

Example 4:

Input: root = [197]
Output: []

Example 5:

Input: root = [31,null,78,null,28]
Output: [78,28]

Constraints:

  • The number of nodes in the tree is in the range [1, 1000].
  • Each node’s value is between [1, 10^6].

Explanation

Traverse the tree, if the tree only has one side of subtree, add that subtree value to the result list, and keep traversing the subtree.

Python Solution

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def getLonelyNodes(self, root: TreeNode) -> List[int]:        
        self.results = []
        
        self.helper(root)
        
        return self.results
    
    def helper(self, root):
        if not root:
            return
        
        if not root.left and not root.right:
            return
        
        if not root.left:
            self.results.append(root.right.val)
            self.helper(root.right)

        elif not root.right:
            self.results.append(root.left.val)            
            self.helper(root.left)
        
        else:
            self.helper(root.left)
            self.helper(root.right)
        
  • Time Complexity: O(N).
  • Space Complexity: O(N).

Leave a Reply

Your email address will not be published.