leetcode -- 102. Binary Tree Level Order Traversal(层次遍历)

题目描述

题目难度：Medium

Given a binary tree, return the level order traversal of its nodes’ values. (ie, from left to right, level by level).

AC代码1

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> resList = new ArrayList<>();
        if(root == null) return resList;
        Queue<TreeNode> queue = new LinkedList<>();
        int start = 0, end = 1;
        queue.offer(root);
        TreeNode node;
        while(!queue.isEmpty()){
            start = 0;
            end = queue.size();
            List<Integer> list = new ArrayList<>();
            while(start < end){
                node = queue.poll();
                list.add(node.val);
                if(node.left != null) queue.offer(node.left);  
                if(node.right != null) queue.offer(node.right);
                start++;
            }
            resList.add(list);
        }
        return resList;
    }
}

AC代码2

class Solution {
    List<List<Integer>> levels =new ArrayList<List<Integer>>();

   public  void helper(TreeNode node , int level){
       // start the current level;
       if(levels.size()==level)
            levels.add(new ArrayList<Integer>());
       //fullfill the current level
       levels.get(level).add(node.val);
       
       //process  child nodes for the levels
       if(node.left!=null)
           helper(node.left, level+1);
       if(node.right!=null)
           helper(node.right, level+1);
       }

    
    public List<List<Integer>> levelOrder(TreeNode root){
        if (root==null) return levels;
        helper(root, 0);
        return levels;
    }
}