Easy Approach with in detail explanation.
Intuition\n Describe your first thoughts on how to solve this problem. \nIn postorder traversal for an n-ary tree, we first traverse all the children of a node 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIn postorder traversal for an n-ary tree, we first traverse all the children of a node before visiting the node itself. The typical order for postorder is: for each node, recursively do a postorder traversal of each of its children, then visit the node.\n\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Use a Stack:\n- We can use a stack to simulate the recursive approach.\n- Start by pushing the root node onto the stack.\n- Pop a node from the stack, and push its value onto a result list (in reverse order).\n- Push all of its children onto the stack (since it\'s postorder, we push leftmost children last).\n2. Reverse the Result:\n- Since the result is collected in reverse postorder, we need to reverse the list before returning it.\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(N), where N is the number of nodes. Each node is visited once.\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nO(N), for the stack and the result list, which in the worst case could contain all nodes.\n\n# Code\n```javascript []\n/**\n * // Definition for a _Node.\n * function _Node(val,children) {\n *    this.val = val;\n *    this.children = children;\n * };\n */\n\n/**\n * @param {_Node|null} root\n * @return {number[]}\n */\nvar postorder = function(root) {\n    if (!root) return [];\n    \n    let stack = [root];\n    let result = [];\n    \n    while (stack.length > 0) {\n        let node = stack.pop();\n        result.push(node.val);\n        \n        // Push the children to the stack in the order they appear\n        for (let child of node.children) {\n            stack.push(child);\n        }\n    }\n    \n    // Reverse the result to get the correct postorder\n    return result.reverse();\n};\n```
['JavaScript']