【LeetCode】102. Binary Tree Level Order Traversal

时间:2019-11-04
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Difficulty: Medium

 More:【目录】LeetCode Java实现

Description

https://leetcode.com/problems/binary-tree-level-order-traversal/

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]
]

Intuition

Using a queue.

Solution

    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> wrapList = new LinkedList<List<Integer>>();
        LinkedList<TreeNode> queue = new LinkedList<>();
        if(root!=null)
            queue.offer(root);
        while(!queue.isEmpty()){
            int num = queue.size();
            List<Integer> subList = new LinkedList<Integer>();
            for(int i=0; i<num; i++){
                TreeNode node = queue.poll();
                subList.add(node.val);
                if(node.left!=null)
                    queue.offer(node.left);
                if(node.right!=null)
                    queue.offer(node.right);
            }
            wrapList.add(subList);  
        }
        return wrapList;
    }

  

Complexity

Time complexity : O(n)

Space complexity : O(n)

 

 More:【目录】LeetCode Java实现

原文地址:https://www.cnblogs.com/yongh/p/11791511.html