Leetcode problem 135 - N-ary Tree Postorder Traversal

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.

/**
 * // Definition for a Node.
 * function Node(val,children) {
 *    this.val = val;
 *    this.children = children;
 * };
 */

/**
 * @param {Node|null} root
 * @return {number[]}
 */
var postorder = function (root) {
  let stack = [];

  const traverse = (node) => {
    if (!node) return;

    for (let child of node.children) {
      traverse(child);
    }
    stack.push(node.val);
  };
  traverse(root);

  return stack;
};

The time complexity of this solution is O(n), where n is the number of nodes in the tree. This is because we are visiting each node exactly once in a depth-first manner.

The space complexity is also O(n) in the worst case, where n is the number of nodes in the tree. This is because we are using a stack to store the nodes in postorder traversal, and in the worst case scenario, the stack can contain all the nodes of the tree.