「力扣」第 102 题:二叉树的层次遍历
传送门:102. 二叉树的层次遍历。
给定一个二叉树,返回其按层次遍历的节点值。 (即逐层地,从左到右访问所有节点)。
例如:
给定二叉树:[3,9,20,null,null,15,7]
,3 / \ 9 20 / \ 15 7
返回其层次遍历结果:
[ [3], [9,20], [15,7] ]
分析:非常标准的层序遍历的做法,使用队列作为辅助的数据结构。
Java 代码:
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;
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;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while (!queue.isEmpty()) {
int size = queue.size();
List<Integer> currentLevel = new ArrayList<>(size);
for (int i = 0; i < size; i++) {
TreeNode head = queue.poll();
currentLevel.add(head.val);
if (head.left != null) {
queue.offer(head.left);
}
if (head.right != null) {
queue.offer(head.right);
}
}
res.add(currentLevel);
}
return res;
}
}
Python 代码:
from typing import List
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
class Solution:
def levelOrder(self, root: TreeNode) -> List[List[int]]:
res = []
if root is None:
return res
queue = [root]
while queue:
size = len(queue)
cur = []
for _ in range(size):
top = queue.pop(0)
cur.append(top.val)
if top.left:
queue.append(top.left)
if top.right:
queue.append(top.right)
res.append(cur)
return res