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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 or DFS.

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 $ans$.

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.

  • /*
// 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;
    }
}


// Solution 2
/*
// 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 {
    private List<List<Integer>> ans = new ArrayList<>();

    public List<List<Integer>> levelOrder(Node root) {
        dfs(root, 0);
        return ans;
    }

    private void dfs(Node root, int i) {
        if (root == null) {
            return;
        }
        if (ans.size() <= i) {
            ans.add(new ArrayList<>());
        }
        ans.get(i++).add(root.val);
        for (Node child : root.children) {
            dfs(child, i);
        }
    }
}


  • /*
// 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 > 0; --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;
    }
};


// Solution 2
/*
// 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;
        function<void(Node*, int i)> dfs = [&](Node* root, int i) {
            if (!root) {
                return;
            }
            if (ans.size() <= i) {
                ans.push_back({});
            }
            ans[i++].push_back(root->val);
            for (auto& child : root->children) {
                dfs(child, i);
            }
        };
        dfs(root, 0);
        return ans;
    }
};


  • """
# 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


# Solution 2
"""
# 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]]:
        def dfs(root, i):
            if root is None:
                return
            if len(ans) <= i:
                ans.append([])
            ans[i].append(root.val)
            for child in root.children:
                dfs(child, i + 1)

        ans = []
        dfs(root, 0)
        return ans


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

func levelOrder(root *Node) [][]int {
	var ans [][]int
	if root == nil {
		return ans
	}
	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 ans
}


// Solution 2
/**
 * Definition for a Node.
 * type Node struct {
 *     Val int
 *     Children []*Node
 * }
 */

func levelOrder(root *Node) (ans [][]int) {
	var dfs func(root *Node, i int)
	dfs = func(root *Node, i int) {
		if root == nil {
			return
		}
		if len(ans) <= i {
			ans = append(ans, []int{})
		}
		ans[i] = append(ans[i], root.Val)
		for _, child := range root.Children {
			dfs(child, i+1)
		}
	}
	dfs(root, 0)
	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 res = [];
    if (root == null) {
        return res;
    }
    const queue = [root];
    while (queue.length !== 0) {
        const n = queue.length;
        const vals = [];
        for (let i = 0; i < n; i++) {
            const { val, children } = queue.shift();
            vals.push(val);
            queue.push(...children);
        }
        res.push(vals);
    }
    return res;
}


// Solution 2
/**
 * 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[][] = [];
    const dfs = (root: Node | null, i: number) => {
        if (root === null) {
            return;
        }
        if (ans.length <= i) {
            ans.push([]);
        }
        const { val, children } = root;
        ans[i++].push(val);
        children.forEach(node => dfs(node, i));
    };
    dfs(root, 0);
    return ans;
}

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