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Formatted question description: https://leetcode.ca/all/589.html

589. N-ary Tree Preorder Traversal (Easy)

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

For example, given a 3-ary tree:

 

 

Return its preorder traversal as: [1,3,5,6,2,4].

 

Note:

Recursive solution is trivial, could you do it iteratively?

Solution 1. Recursive

// OJ: https://leetcode.com/problems/n-ary-tree-preorder-traversal/
// Time: O(N)
// Space: O(logN)
class Solution {
private:
    vector<int> ans;
    void rec(Node *root) {
        if (!root) return;
        ans.push_back(root->val);
        for (auto ch : root->children) rec(ch);
    }
public:
    vector<int> preorder(Node* root) {
        rec(root);
        return ans;
    }
};

Solution 2. Iterative

// OJ: https://leetcode.com/problems/n-ary-tree-preorder-traversal/
// Time: O(N)
// Space: O(logN)
class Solution {
public:
    vector<int> preorder(Node* root) {
        if (!root) return {};
        vector<int> ans;
        stack<Node*> s;
        s.push(root);
        while (s.size()) {
            root = s.top();
            s.pop();
            ans.push_back(root->val);
            for (int i = root->children.size() - 1; i >= 0; --i) {
                if (root->children[i]) s.push(root->children[i]);
            }
        }
        return ans;
    }
};

• Java
• C++
• Python
• Go
• TypeScript

• /*
  // 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<Integer> preorder(Node root) {
          List<Integer> preorder = new ArrayList<Integer>();
          if (root == null)
              return preorder;
          Stack<Node> stack = new Stack<Node>();
          stack.push(root);
          while (!stack.isEmpty()) {
              Node node = stack.pop();
              preorder.add(node.val);
              List<Node> children = node.children;
              int numOfChildren = children.size();
              for (int i = numOfChildren - 1; i >= 0; i--)
                  stack.push(children.get(i));
          }
          return preorder;
      }
  }
  
  ############
  
  /*
  // 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<Integer> preorder(Node root) {
          if (root == null) {
              return Collections.emptyList();
          }
          List<Integer> ans = new ArrayList<>();
          Deque<Node> stk = new ArrayDeque<>();
          stk.push(root);
          while (!stk.isEmpty()) {
              Node node = stk.pop();
              ans.add(node.val);
              List<Node> children = node.children;
              for (int i = children.size() - 1; i >= 0; --i) {
                  stk.push(children.get(i));
              }
          }
          return ans;
      }
  }
  

• // OJ: https://leetcode.com/problems/n-ary-tree-preorder-traversal/
  // Time: O(N)
  // Space: O(logN)
  class Solution {
  private:
      vector<int> ans;
      void rec(Node *root) {
          if (!root) return;
          ans.push_back(root->val);
          for (auto ch : root->children) rec(ch);
      }
  public:
      vector<int> preorder(Node* root) {
          rec(root);
          return ans;
      }
  };
  

• """
  # Definition for a Node.
  class Node:
      def __init__(self, val=None, children=None):
          self.val = val
          self.children = children
  """
  
  
  class Solution:
      def preorder(self, root: 'Node') -> List[int]:
          ans = []
          if root is None:
              return ans
          stk = [root]
          while stk:
              node = stk.pop()
              ans.append(node.val)
              for child in node.children[::-1]:
                  stk.append(child)
          return ans
  
  ############
  
  """
  # Definition for a Node.
  class Node(object):
      def __init__(self, val, children):
          self.val = val
          self.children = children
  """
  class Solution(object):
      def preorder(self, root):
          """
          :type root: Node
          :rtype: List[int]
          """
          res = []
          if not root:
              return res
          res.append(root.val)
          for child in root.children:
              res.extend(self.preorder(child))
          return res
  

• /**
   * Definition for a Node.
   * type Node struct {
   *     Val int
   *     Children []*Node
   * }
   */
  
  func preorder(root *Node) []int {
  	var ans []int
  	if root == nil {
  		return ans
  	}
  	stk := []*Node{root}
  	for len(stk) > 0 {
  		node := stk[len(stk)-1]
  		ans = append(ans, node.Val)
  		stk = stk[:len(stk)-1]
  		children := node.Children
  		for i := len(children) - 1; i >= 0; i-- {
  			stk = append(stk, children[i])
  		}
  	}
  	return ans
  }
  

• /**
   * Definition for node.
   * class Node {
   *     val: number
   *     children: Node[]
   *     constructor(val?: number) {
   *         this.val = (val===undefined ? 0 : val)
   *         this.children = []
   *     }
   * }
   */
  
  function preorder(root: Node | null): number[] {
      const ans = [];
      const dfs = (root: Node | null) => {
          if (root == null) {
              return;
          }
          ans.push(root.val);
          for (const node of root.children) {
              dfs(node);
          }
      };
      dfs(root);
      return ans;
  }
  
  

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