Question

429. N-ary Tree Level Order Traversal

中文文档

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:

1. Create an empty list $t$ to store the values of the current level nodes.
2. For each node in the queue, add its value to $t$ and add its child nodes to the queue.
3. 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.

"""
# 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 $ans$.

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:
  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.
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;
    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.
 * 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 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;
}