LeetCode 589. N叉树的前序遍历(N-ary Tree Preorder Traversal)

589. N叉树的前序遍历

589. N-ary Tree Preorder Traversal

LeetCode589. N-ary Tree Preorder Traversal

题目描述

给定一个 N 叉树,返回其节点值的前序遍历。

例如,给定一个 3 叉树 :

LeetCode 589. N叉树的前序遍历(N-ary Tree Preorder Traversal)

返回其前序遍历: [1,3,5,6,2,4]。

说明: 递归法很简单,你可以使用迭代法完成此题吗?

Java 实现

Iterative Solution

import java.util.LinkedList;
import java.util.List;
import java.util.Stack; class Node {
public int val;
public List<Node> children; public Node() {
} public Node(int _val, List<Node> _children) {
val = _val;
children = _children;
}
} class Solution {
public List<Integer> preorder(Node root) {
List<Integer> result = new LinkedList<>();
if (root == null) {
return result;
}
Stack<Node> stack = new Stack<>();
stack.push(root);
while (!stack.isEmpty()) {
Node node = stack.pop();
if (node.children != null) {
for (int i = node.children.size() - 1; i >= 0; i--) {
stack.push(node.children.get(i));
}
}
result.add(node.val);
}
return result;
}
}

Recursive Solution

import java.util.LinkedList;
import java.util.List; class Node {
public int val;
public List<Node> children; public Node() {
} public Node(int _val, List<Node> _children) {
val = _val;
children = _children;
}
} class Solution {
List<Integer> result = new LinkedList<>(); public List<Integer> preorder(Node root) {
if (root == null) {
return result;
}
result.add(root.val);
List<Node> node = root.children;
for (int i = 0; i < node.size(); i++) {
preorder(node.get(i));
}
return result;
}
}

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