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).