【树】C003_N叉树的前序遍历（递归 | 迭代）

一、题目描述

Given an n-ary tree, return the preorder 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?

Input: root = [1,null,3,2,4,null,5,6]
Output: [1,3,5,6,2,4]

public class Node {

  public int val;
  public List<Node> children;

  public Node() {}

  public Node(int _val) {
    val = _val;
  }

  public Node(int _val, List<Node> _children) {
    val = _val;
    children = _children;
  }
}

二、题解

方法一：递归

联想二叉树的前序遍历，N 叉树的遍历过程类似，类似先遍历根节点 root，然后依次递归遍历每个子树 node child : root.child.

LinkedList<Integer> resList = null;
public List<Integer> preorder(Node root) {
  resList = new LinkedList<>();
  dfs(root);
  return resList;
}

private void dfs(Node root) {
  if (root == null)
    return;
  resList.add(root.val);
  for (Node child : root.children) {
    dfs(child);
  }
}

复杂度分析

• 时间复杂度： $O(N)$，
• 空间复杂度： $O(N)$，

---

方法二：迭代

解释一下为什么从后往前遍历：因为遍历是从左到右的，所以先得把孩子从右往左压栈，取出结点时的顺序才是从左到右的顺序。

private void dfs(Node root) {
  Stack<Node> stack = new Stack<>();
  stack.push(root);
  while (!stack.isEmpty()) {
    Node node = stack.pop();
    resList.add(node.val);
    //if (root.children == null)
    //  continue;
    for(int i = node.children.size()-1; i >= 0; i--) {
      stack.add(node.children.get(i));
    }  
  }
}

复杂度分析

• 时间复杂度： $O(N)$，
• 空间复杂度： $O(N)$，