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  • [LeetCode] 102. Binary Tree Level Order Traversal Java

    题目:

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

    For example:
    Given binary tree [3,9,20,null,null,15,7],

        3
       / 
      9  20
        /  
       15   7

    return its level order traversal as:

    [
      [3],
      [9,20],
      [15,7]
    ]
    

     题意及分析:宽度遍历一棵树。用队列保存中间值即可。

    代码:

    /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    public class Solution {
        public List<List<Integer>> levelOrder(TreeNode root) {
            List<List<Integer>> res = new ArrayList<>();
            if(root==null) return res;
            List<Integer> list = new ArrayList<>();
            Queue<TreeNode> queue=new LinkedList<>();
            queue.offer(root);
            Queue<TreeNode> queue1=new LinkedList<>();    //这里也可以用size来界定
            while(!queue.isEmpty()||!queue1.isEmpty()){
                queue1.clear();
                while(!queue.isEmpty()){        //遍历一层
                    TreeNode now = queue.poll();
                    list.add(now.val);
                    if(now.left!=null)
                        queue1.offer(now.left);
                    if(now.right!=null) 
                        queue1.offer(now.right);
                }
                queue=new LinkedList<>(queue1);       //进行下一层的遍历
                res.add(new ArrayList<>(list));
                list.clear();       //清空
            }
            return res;
        }
    }
    

      


     
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  • 原文地址:https://www.cnblogs.com/271934Liao/p/7197331.html
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