leetcode(NOWCODER)---binary-tree-preorder-traversal

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时间限制：1秒 空间限制：32768K 热度指数：23954
本题知识点： 树 leetcode
算法知识视频讲解
题目描述

Given a binary tree, return the preorder traversal of its nodes’ values.

For example:
Given binary tree{1,#,2,3},

1

2
/
3

return[1,2,3].

Note: Recursive solution is trivial, could you do it iteratively?

思路：https://blog.csdn.net/ccnuacmhdu/article/details/85227092

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
import java.util.*;
public class Solution {
    private ArrayList<Integer> list = new ArrayList<Integer>();
    public ArrayList<Integer> preorderTraversal(TreeNode root) {
        //这道题明显类似于上一题后序遍历的解法，后序遍历用栈:
        //https://blog.csdn.net/ccnuacmhdu/article/details/85227092
        //这道题依然是用栈！！细想下，不能用队列！！递归与栈本一家啊，两道非递归遍历题感触深
        if(root == null){
            return list;
        }
        ArrayDeque<TreeNode> stack = new ArrayDeque<TreeNode>();
        stack.addFirst(root);
        while(stack.isEmpty() == false){
            TreeNode node = stack.removeFirst();
            list.add(node.val);
            if(node.right != null){
                stack.addFirst(node.right);
            }
            if(node.left != null){
                stack.addFirst(node.left);
            }
        }
        return list;
    }
}