429. N-ary Tree Level Order Traversal

Description

Given an n-ary tree, return the level order traversal of its nodes' values.

Nary-Tree input serialization is represented in their level order traversal, each group of children is separated by the null value (See examples).

Example 1:

Input: root = [1,null,3,2,4,null,5,6]
Output: [[1],[3,2,4],[5,6]]

Example 2:

Input: root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]
Output: [[1],[2,3,4,5],[6,7,8,9,10],[11,12,13],[14]]

Constraints:

The height of the n-ary tree is less than or equal to 1000
The total number of nodes is between [0, 10⁴]

Solutions

Solution 1: BFS

First, we check if the root node is null. If it is, we return an empty list directly.

Otherwise, we create a queue \(q\) and initially add the root node to the queue.

When the queue is not empty, we loop through the following operations:

Create an empty list \(t\) to store the values of the current level nodes.
For each node in the queue, add its value to \(t\) and add its child nodes to the queue.
Add \(t\) to the result list \(ans\).

Finally, return the result list \(ans\).

The time complexity is \(O(n)\), and the space complexity is \(O(n)\). Here, \(n\) is the number of nodes in the N-ary tree.

Python3JavaC++GoTypeScript

"""
# Definition for a Node.
class Node:
    def __init__(self, val=None, children=None):
        self.val = val
        self.children = children
"""


class Solution:
    def levelOrder(self, root: 'Node') -> List[List[int]]:
        ans = []
        if root is None:
            return ans
        q = deque([root])
        while q:
            t = []
            for _ in range(len(q)):
                root = q.popleft()
                t.append(root.val)
                q.extend(root.children)
            ans.append(t)
        return ans

/*
// Definition for a Node.
class Node {
    public int val;
    public List<Node> children;

    public Node() {}

    public Node(int _val) {
        val = _val;
    }

    public Node(int _val, List<Node> _children) {
        val = _val;
        children = _children;
    }
};
*/

class Solution {
    public List<List<Integer>> levelOrder(Node root) {
        List<List<Integer>> ans = new ArrayList<>();
        if (root == null) {
            return ans;
        }
        Deque<Node> q = new ArrayDeque<>();
        q.offer(root);
        while (!q.isEmpty()) {
            List<Integer> t = new ArrayList<>();
            for (int n = q.size(); n > 0; --n) {
                root = q.poll();
                t.add(root.val);
                q.addAll(root.children);
            }
            ans.add(t);
        }
        return ans;
    }
}

/*
// Definition for a Node.
class Node {
public:
    int val;
    vector<Node*> children;

    Node() {}

    Node(int _val) {
        val = _val;
    }

    Node(int _val, vector<Node*> _children) {
        val = _val;
        children = _children;
    }
};
*/

class Solution {
public:
    vector<vector<int>> levelOrder(Node* root) {
        vector<vector<int>> ans;
        if (!root) {
            return ans;
        }
        queue<Node*> q{{root}};
        while (!q.empty()) {
            vector<int> t;
            for (int n = q.size(); n; --n) {
                root = q.front();
                q.pop();
                t.push_back(root->val);
                for (auto& child : root->children) {
                    q.push(child);
                }
            }
            ans.push_back(t);
        }
        return ans;
    }
};

/**
 * Definition for a Node.
 * type Node struct {
 *     Val int
 *     Children []*Node
 * }
 */

func levelOrder(root *Node) (ans [][]int) {
    if root == nil {
        return
    }
    q := []*Node{root}
    for len(q) > 0 {
        var t []int
        for n := len(q); n > 0; n-- {
            root = q[0]
            q = q[1:]
            t = append(t, root.Val)
            for _, child := range root.Children {
                q = append(q, child)
            }
        }
        ans = append(ans, t)
    }
    return
}

/**
 * Definition for node.
 * class Node {
 *     val: number
 *     children: Node[]
 *     constructor(val?: number) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.children = []
 *     }
 * }
 */

function levelOrder(root: Node | null): number[][] {
    const ans: number[][] = [];
    if (!root) {
        return ans;
    }
    const q: Node[] = [root];
    while (q.length) {
        const qq: Node[] = [];
        const t: number[] = [];
        for (const { val, children } of q) {
            qq.push(...children);
            t.push(val);
        }
        ans.push(t);
        q.splice(0, q.length, ...qq);
    }
    return ans;
}

Solution 2: DFS

We can use the Depth-First Search method to traverse the entire tree.

We define a helper function \(dfs(root, i)\), where \(root\) represents the current node, and \(i\) represents the current level.

In the \(dfs\) function, we first check if \(root\) is null. If it is, we return directly.

Otherwise, we check if the length of \(ans\) is less than or equal to \(i\). If it is, it means that the current level has not been added to \(ans\) yet, so we need to add an empty list first. Then we add the value of \(root\) to \(ans[i]\).

Next, we traverse all child nodes of \(root\). For each child node, we call \(dfs(child, i + 1)\).

In the main function, we call \(dfs(root, 0)\) and return \(ans\).