Leetcode skimming: binary tree 04 (sequence traversal of binary tree)

102. Sequence traversal of binary tree

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Give you a binary tree , Please return to its button Sequence traversal The resulting node value . （ That is, layer by layer , Access all nodes from left to right ）.

At first glance, it looks complicated , Do it carefully , One move per second .

package com.programmercarl.tree;

import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Deque;
import java.util.List;

/** * @ClassName LevelOrder * @Descriotion TODO * @Author nitaotao * @Date 2022/7/3 11:55 * @Version 1.0 * https://leetcode.cn/problems/binary-tree-level-order-traversal/ * 102.  Sequence traversal of binary tree  **/
public class LevelOrder {
    
    public List<List<Integer>> levelOrder(TreeNode root) {
    
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        Deque<TreeNode> deque = new ArrayDeque();
        if (root == null) {
    
            return result;
        }
        // The team 
        deque.offer(root);
        while (!deque.isEmpty()) {
    
            int size=deque.size();
            List<Integer> floor = new ArrayList();
            while (size > 0) {
    
                // Traverse one layer at a time 
                TreeNode node = deque.poll();
                floor.add(node.val);
                if (node.left != null) {
    
                    deque.offer(node.left);
                }
                if (node.right != null) {
    
                    deque.offer(node.right);
                }
                size--;
            }
            result.add(floor);
        }
        return result;
    }
}

Sequence traversal , Big list
Bao Xiao list . Every little list For the first floor .
Layer by layer , Record the length of the current queue , That is, how many elements are there on the upper layer , How many times will it pop up this time , Add new elements to the end of the queue .

 Look at the solution , Mine belongs to breadth first traversal , And depth first traversal .

/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * this.left = left; * this.right = right; * } * } */
class Solution {
    
    public List<List<Integer>> resList = new ArrayList<List<Integer>>();
    public List<List<Integer>> levelOrder(TreeNode root) {
    
        checkFun01(root, 0);
        return resList;
    }

    public void checkFun01(TreeNode node, Integer deep) {
    
        if (node == null) {
    
            return;
        }
        deep++;
        if (resList.size() < deep) {
    
            // As the hierarchy increases ,list Of Item Also increase , utilize List Index value of the hierarchy node 
            List<Integer> item = new ArrayList<>();
            resList.add(item);
        }
        resList.get(deep - 1).add(node.val);
        checkFun01(node.left, deep);
        checkFun01(node.right, deep);
    }
}

I don't understand what I see , It's a bit like preparing one first List , Then depth first traversal , Put the index of the result according to the sequence traversal .