Given 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).
Follow up:
Recursive solution is trivial, could you do it iteratively?
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:
1000[0, 10^4]class Solution { List<Integer> list = new ArrayList<>(); public List<Integer> postorder(Node root) { if (root == null) return list; for(Node node: root.children) postorder(node); list.add(root.val); return list; } }
class Solution { public List<Integer> postorder(Node root) { if(root == null) return Collections.emptyList(); LinkedList<Integer> list = new LinkedList<>(); Stack<Node> stack = new Stack<>(); stack.push(root); while(!stack.isEmpty()){ Node n = stack.pop(); if(n.children != null){ for(Node node: n.children){ stack.push(node); } } list.addFirst(n.val); } return list; } }
590. N-ary Tree Postorder Traversal
原文：https://www.cnblogs.com/wentiliangkaihua/p/13184897.html