LeetCode之Binary Tree Level Order Traversal 层序遍历二叉树

2024-05-12 05:32:14

题目描述：

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,#,#,15,7},
    3/ \9  20/  \15   7
return its level order traversal as:
[[3],[9,20],[15,7]
]

题目解答：

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/**

* 实质是二叉树的广度优先搜索

* 利用一个辅助队列保存被访问的当前节点的左右孩子，

* 调整进出队列的顺序以实现层序遍历

*/

public List<List<Integer>> levelOrder(TreeNode root) {

List<List<Integer>> result=new ArrayList<>();

if(root==null){

return result;

}

/**

* 一般来说，队列的实现选择LinkedList即可

* 队列保存节点就不用找什么变量了

*/

Queue<TreeNode> queue=new LinkedList<>();

queue.offer(root);

//首先把头结点的值保存到结果集，然后把左右子节点分别进入队列

while(!queue.isEmpty()){//需要使用isEmptyy判断，不能使用null

List<Integer> temp=new ArrayList<>();

/**

* error!这里对进行循环的过程，队列长度是在不断变化的

* size需要等于队列出队前的长度

*/

int size =queue.size();

for(int i=0;i<size;i++){

TreeNode node=queue.poll();

temp.add(node.val);

if(node.left!=null)

queue.offer(node.left);

if(node.right!=null)

queue.offer(node.right);

}

//当前队列全部poll，到这里已经完成了一层的遍历

result.add(temp);

}

return result;

}

　　

本文转自邴越博客园博客，原文链接：http://www.cnblogs.com/binyue/p/4250267.html，如需转载请自行联系原作者

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