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,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]
解题思路:层次遍历一棵二叉树,采用先序遍历的方式,用一个队列的数据结构存放遍历到的节点,扩展节点时再依次出队
class Solution {
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) {
val = x;
}
}
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> list = new LinkedList<List<Integer>>();
if(root == null) {
return list;
}
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.add(root);
while(!queue.isEmpty()) {
List<Integer> sublist = new LinkedList<Integer>();
//注意此处size要和queue.size单独分开,因为在循环体中还会动态的向队列中添加节点
//再次遍历时,此时queue.size是变化的
int size = queue.size();
for(int i = 0; i < size; i++) {
TreeNode currentNode = queue.poll();
sublist.add(currentNode.val);
if(currentNode.left!=null) {
queue.add(currentNode.left);
}
if(currentNode.right!=null) {
queue.add(currentNode.right);
}
}
list.add(sublist);
}
return list;
}
}