Formatted question description: https://leetcode.ca/all/1469.html
1469. Find All the Lonely Nodes
Level
Easy
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]
.
Solution
Create a list to store the values of all lonely nodes. If root
is null
, return the list directly.
Do breadth first search. For each node, check its two children. For each non-empty child, offer the child to the queue for the next step’s search. If exactly one child is non-empty, then add the non-empty child’s value into the list.
Finally, return the list.
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<Integer> getLonelyNodes(TreeNode root) {
List<Integer> lonelyNodes = new ArrayList<Integer>();
if (root == null)
return lonelyNodes;
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.offer(root);
while (!queue.isEmpty()) {
TreeNode node = queue.poll();
TreeNode left = node.left, right = node.right;
if (left != null && right != null) {
queue.offer(left);
queue.offer(right);
} else if (left != null) {
lonelyNodes.add(left.val);
queue.offer(left);
} else if (right != null) {
lonelyNodes.add(right.val);
queue.offer(right);
}
}
return lonelyNodes;
}
}