From 8364dcd0f181e6152f00b6f6bf415f296b451e0b Mon Sep 17 00:00:00 2001
From: jwcen <jvvcen@gamil.com>
Date: Wed, 7 Dec 2022 19:43:34 +0800
Subject: [PATCH] =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=A0=91=E7=9A=84=E5=B1=82?=
 =?UTF-8?q?=E5=BA=8F=E9=81=8D=E5=8E=86=EF=BC=9A=E4=BD=BF=E7=94=A8=E5=88=87?=
 =?UTF-8?q?=E7=89=87=E6=A8=A1=E6=8B=9F=E9=98=9F=E5=88=97=EF=BC=8C=E6=98=93?=
 =?UTF-8?q?=E7=90=86=E8=A7=A3?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 problems/0102.二叉树的层序遍历.md | 30 +++++++++++++++++++++++
 1 file changed, 30 insertions(+)

diff --git a/problems/0102.二叉树的层序遍历.md b/problems/0102.二叉树的层序遍历.md
index 9985971f..bc78cb36 100644
--- a/problems/0102.二叉树的层序遍历.md
+++ b/problems/0102.二叉树的层序遍历.md
@@ -271,6 +271,36 @@ func levelOrder(root *TreeNode) [][]int {
     
     return res
 }
+
+/**
+102. 二叉树的层序遍历：使用切片模拟队列，易理解
+ */
+func levelOrder(root *TreeNode) (res [][]int) {
+    if root == nil {
+        return 
+    }
+
+    curLevel := []*TreeNode{root}  // 存放当前层节点
+    for len(curLevel) > 0 {
+        nextLevel := []*TreeNode{}  // 准备通过当前层生成下一层
+        vals := []int{}
+
+        for _, node := range curLevel {
+            vals = append(vals, node.Val) // 收集当前层的值
+            // 收集下一层的节点
+            if node.Left != nil {
+                nextLevel = append(nextLevel, node.Left)
+            }
+            if node.Right != nil {
+                nextLevel = append(nextLevel, node.Right)
+            }
+        }
+        res = append(res, vals)
+        curLevel = nextLevel // 将下一层变成当前层
+    }
+
+    return
+}
 ```
 
 javascript代码：