N-ary Tree Postorder Traversal LeetCode Solution

Problem – N-ary Tree Postorder Traversal LeetCode Solution

Given the root of an n-ary tree, return the postorder 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: [5,6,3,2,4,1]

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: [2,6,14,11,7,3,12,8,4,13,9,10,5,1]

Constraints:

• The number of nodes in the tree is in the range [0, 104].
• 0 <= Node.val <= 104
• The height of the n-ary tree is less than or equal to 1000.

N-ary Tree Postorder Traversal LeetCode Solution in Java

class Solution {
    public List<Integer> postorder(Node root) {
        List<Integer> list = new ArrayList<>();
        if (root == null) return list;
        
        Stack<Node> stack = new Stack<>();
        stack.add(root);
        
        while(!stack.isEmpty()) {
            root = stack.pop();
            list.add(root.val);
            for(Node node: root.children)
                stack.add(node);
        }
        Collections.reverse(list);
        return list;
    }
}

N-ary Tree Postorder Traversal LeetCode Solution in C++

vector<int> postorder(Node* root) {
    if(root==NULL) return {};
    vector<int> res;
    stack<Node*> stk;
    stk.push(root);
    while(!stk.empty())
    {
        Node* temp=stk.top();
        stk.pop();
        for(int i=0;i<temp->children.size();i++) stk.push(temp->children[i]);
        res.push_back(temp->val);
    }
    reverse(res.begin(), res.end());
    return res;
}

N-ary Tree Postorder Traversal LeetCode Solution in Python

def postorder(self, root):
        """
        :type root: Node
        :rtype: List[int]
        """
        res = []
        if root == None: return res

        def recursion(root, res):
            for child in root.children:
                recursion(child, res)
            res.append(root.val)

        recursion(root, res)
        return res

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