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1469. Find All The Lonely Nodes

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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].

Solutions

Python3

# 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]:
        def traverse(root):
            if root is None:
                return
            if root.left is None and root.right is not None:
                self.res.append(root.right.val)
            if root.left is not None and root.right is None:
                self.res.append(root.left.val)
            traverse(root.left)
            traverse(root.right)
        self.res = []
        traverse(root)
        return self.res

Java

/**
 * 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 {
    private List<Integer> res;

    public List<Integer> getLonelyNodes(TreeNode root) {
        res = new ArrayList<>();
        traverse(root);
        return res;
    }

    private void traverse(TreeNode root) {
        if (root == null) {
            return;
        }
        if (root.left == null && root.right != null) {
            res.add(root.right.val);
        }
        if (root.left != null && root.right == null) {
            res.add(root.left.val);
        }
        traverse(root.left);
        traverse(root.right);
    }
}

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