Question

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.

 

Constraints:

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

Solutions

Solution 1

# 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.
 * 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.
 * 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
}