LeetCode - 589. N叉树的前序遍历 Swift &
2020-09-10 本文已影响0人
huxq_coder

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 :

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?
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/n-ary-tree-preorder-traversal
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算法
Java
- 递归
class Solution {
List<Integer> list;
Stack<Node> stack;
/** 递归 */
public List<Integer> preorder(Node root) {
list = new ArrayList<>();
if (root == null) {
return list;
}
helper(root);
return list;
}
public void helper(Node node) {
list.add(node.val);
if (node.children.size() > 0) {
for (Node child : node.children) {
helper(child);
}
}
}
}
- 迭代
/** 迭代 */
public List<Integer> preorder(Node root) {
List<Integer> list = new ArrayList<>();
Stack<Node> stack = new Stack<>();
if (root == null) {
return list;
}
stack.push(root);
while (!stack.isEmpty()) {
Node current = stack.pop();
list.add(current.val);
if (current.children.size() > 0) {
Collections.reverse(current.children);
for (Node child : current.children) {
stack.push(child);
}
}
}
return list;
}
swift
树结构
// Definition for a Node.
public class Node {
public var val: Int
public var children: [Node]
public init(_ val: Int) {
self.val = val
self.children = []
}
}
- 递归
/**
递归
*/
func preorder(_ root: Node?) -> [Int] {
var result = [Int]()
if root == nil {
return result
}
helper(root!, &result)
return result
}
func helper(_ current: Node, _ result: inout [Int]) {
result.append(current.val)
if !current.children.isEmpty {
for child in current.children {
helper(child, &result)
}
}
}
- 迭代
/**
迭代
*/
func preorder(_ root: Node?) -> [Int] {
var result = [Int]()
var stack = [Node]()
if root == nil {
return result
}
stack.append(root!)
while !stack.isEmpty {
let current = stack.popLast()
result.append(current!.val)
if !current!.children.isEmpty {
for child in current!.children.reversed() {
stack.append(child)
}
}
}
return result
}
GitHub:https://github.com/huxq-coder/LeetCode
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