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 $ans$.