1469. Find All The Lonely Nodes

LeetCode 1469. Find All The Lonely Nodes

Description

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].

Tags

Tree

Solution

We collect node values from the views of their parents. In the DFS function, we do not consider any null nodes or leaf nodes since they have no children. Then, if the current node has only one child, we append the value of that child to the result array. At last, perform DFS on both left and right children. Because we only collect the value of input node's child, we can start DFS from root, which will not be collected.

Complexity

• Time complexity: $O(n)$

• Space complexity: $O(1)$

Code

func getLonelyNodes(root *TreeNode) []int {
	var ans []int
	var dfs func(node *TreeNode)
	dfs = func(node *TreeNode) {
		if node == nil || node.Left == nil && node.Right == nil {
			return
		}
		if node.Left == nil {
			ans = append(ans, node.Right.Val)
		}
		if node.Right == nil {
			ans = append(ans, node.Left.Val)
		}
		dfs(node.Left)
		dfs(node.Right)
	}
	dfs(root)
	return ans
}