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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.

 

Constraints:

• The number of nodes in the tree is in the range [1, 1000].
• 1 <= Node.val <= 106

Solutions

• Java
• C++
• Python
• Go

• /**
   * 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> ans = new ArrayList<>();
  
      public List<Integer> getLonelyNodes(TreeNode root) {
          dfs(root);
          return ans;
      }
  
      private void dfs(TreeNode root) {
          if (root == null || (root.left == null && root.right == null)) {
              return;
          }
          if (root.left == null) {
              ans.add(root.right.val);
          }
          if (root.right == null) {
              ans.add(root.left.val);
          }
          dfs(root.left);
          dfs(root.right);
      }
  }
  

• /**
   * Definition for a binary tree node.
   * struct TreeNode {
   *     int val;
   *     TreeNode *left;
   *     TreeNode *right;
   *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
   *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
   *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
   * };
   */
  class Solution {
  public:
      vector<int> getLonelyNodes(TreeNode* root) {
          vector<int> ans;
          function<void(TreeNode * root)> dfs;
          dfs = [&](TreeNode* root) {
              if (!root || (!root->left && !root->right)) return;
              if (!root->left) ans.push_back(root->right->val);
              if (!root->right) ans.push_back(root->left->val);
              dfs(root->left);
              dfs(root->right);
          };
          dfs(root);
          return ans;
      }
  };
  

• # 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: Optional[TreeNode]) -> List[int]:
          def dfs(root):
              if root is None or (root.left is None and root.right is None):
                  return
              if root.left is None:
                  ans.append(root.right.val)
              if root.right is None:
                  ans.append(root.left.val)
              dfs(root.left)
              dfs(root.right)
  
          ans = []
          dfs(root)
          return ans
  
  

• /**
   * Definition for a binary tree node.
   * type TreeNode struct {
   *     Val int
   *     Left *TreeNode
   *     Right *TreeNode
   * }
   */
  func getLonelyNodes(root *TreeNode) []int {
  	ans := []int{}
  	var dfs func(*TreeNode)
  	dfs = func(root *TreeNode) {
  		if root == nil || (root.Left == nil && root.Right == nil) {
  			return
  		}
  		if root.Left == nil {
  			ans = append(ans, root.Right.Val)
  		}
  		if root.Right == nil {
  			ans = append(ans, root.Left.Val)
  		}
  		dfs(root.Left)
  		dfs(root.Right)
  	}
  	dfs(root)
  	return ans
  }
  

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