建设银行手机银行网站登录,静态网站模板古典,wordpress 菜单消失,安庆网站建设专590. N 叉树的后序遍历
给定一个 n 叉树的根节点 root #xff0c;返回 其节点值的 后序遍历 。
n 叉树 在输入中按层序遍历进行序列化表示#xff0c;每组子节点由空值 null 分隔#xff08;请参见示例#xff09;。 示例 1#xff1a; 输入#xff1a;root [1,null,…590. N 叉树的后序遍历
给定一个 n 叉树的根节点 root 返回 其节点值的 后序遍历 。
n 叉树 在输入中按层序遍历进行序列化表示每组子节点由空值 null 分隔请参见示例。 示例 1 输入root [1,null,3,2,4,null,5,6]
输出[5,6,3,2,4,1]示例 2 输入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]
输出[2,6,14,11,7,3,12,8,4,13,9,10,5,1]提示
节点总数在范围 [0, 104] 内0 Node.val 104n 叉树的高度小于或等于 1000 进阶递归法很简单你可以使用迭代法完成此题吗?
解法思路 1、递归Recursion 2、迭代Iterator效率低 法一
/*
// Definition for a Node.
class Node {public int val;public ListNode children;public Node() {}public Node(int _val) {val _val;}public Node(int _val, ListNode _children) {val _val;children _children;}
};
*/class Solution {public ListInteger postorder(Node root) {// Recursion// Time: O(n), n 为节点数// Space: O(n)ListInteger res new ArrayList();helper(root, res);return res;}private void helper(Node root, ListInteger res) {if (root null) return;for (Node child : root.children) {helper(child, res);}res.add(root.val);}
} 法二
/*
// Definition for a Node.
class Node {public int val;public ListNode children;public Node() {}public Node(int _val) {val _val;}public Node(int _val, ListNode _children) {val _val;children _children;}
};
*/class Solution {public ListInteger postorder(Node root) {// Iterator// Time: O(n), n 为节点数// Space: O(n)ListInteger res new ArrayList();if (root null) return res;MapNode, Integer map new HashMapNode, Integer();DequeNode stack new ArrayDequeNode();Node node root;while (node ! null || !stack.isEmpty()) {while (node ! null) {stack.addFirst(node);ListNode children node.children;if (!children.isEmpty()) {map.put(node, 0);node children.get(0);} else {node null;}}node stack.peek();int idx map.getOrDefault(node, -1) 1;ListNode children node.children;if (!children.isEmpty() children.size() idx) {map.put(node, idx);node children.get(idx);} else {res.add(node.val);stack.removeFirst();map.remove(node);node null;}}return res;}
} 优化迭代
/*
// Definition for a Node.
class Node {public int val;public ListNode children;public Node() {}public Node(int _val) {val _val;}public Node(int _val, ListNode _children) {val _val;children _children;}
};
*/class Solution {public ListInteger postorder(Node root) {// Optimize Iterator// Time: O(n), n 为节点数// Space: O(n)ListInteger res new ArrayList();if (root null) return res;DequeNode stack new ArrayDeque();SetNode visited new HashSetNode();stack.addFirst(root);while (!stack.isEmpty()) {Node node stack.peek();if (node.children.isEmpty() || visited.contains(node)) {stack.removeFirst();res.add(node.val);continue;}for (int i node.children.size() - 1; i 0; --i) {stack.addFirst(node.children.get(i));}visited.add(node);}return res;}
}
