README_EN.md

comments	difficulty	edit_url	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/1400-1499/1469.Find%20All%20The%20Lonely%20Nodes/README_EN.md	Tree Depth-First Search Breadth-First Search Binary Tree

1469. Find All The Lonely Nodes 🔒

中文文档

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: DFS

We can use Depth-First Search (DFS) to traverse the entire tree. We design a function $\textit{dfs}$, which traverses each node in the tree. If the current node is a lone child, we add its value to the answer array. The execution process of the function $\textit{dfs}$ is as follows:

1. If the current node is null, or the current node is a leaf node (i.e., both the left and right children of the current node are null), then return directly.
2. If the left child of the current node is null, then the right child of the current node is a lone child, and we add its value to the answer array.
3. If the right child of the current node is null, then the left child of the current node is a lone child, and we add its value to the answer array.
4. Recursively traverse the left and right children of the current node.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes in the binary tree.

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: Optional[TreeNode]) -> List[int]:
        def dfs(root: Optional[TreeNode]):
            if root is None or root.left == root.right:
                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

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> ans = new ArrayList<>();

    public List<Integer> getLonelyNodes(TreeNode root) {
        dfs(root);
        return ans;
    }

    private void dfs(TreeNode root) {
        if (root == null || (root.left == root.right)) {
            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);
    }
}

C++

/**
 * 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;
        auto dfs = [&](this auto&& 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;
    }
};

Go

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

TypeScript

/**
 * 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;
}