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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

  • /**
 * 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
}

  • /**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function getLonelyNodes(root: TreeNode | null): number[] {
    const ans: number[] = [];
    const dfs = (root: TreeNode | null) => {
        if (!root || root.left === root.right) {
            return;
        }
        if (!root.left) {
            ans.push(root.right.val);
        }
        if (!root.right) {
            ans.push(root.left.val);
        }
        dfs(root.left);
        dfs(root.right);
    };
    dfs(root);
    return ans;
}

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