add: 二叉树的层序遍历 II

README.md

@@ -359,11 +359,16 @@ LeetCode 与 LintCode 解题记录。此为个人练习仓库，代码中对重
   - LeetCode 572. 另一个树的子树 <https://leetcode-cn.com/problems/subtree-of-another-tree/>
   - LintCode 1165. 另一个树的子树 <https://www.lintcode.com/problem/subtree-of-another-tree/description>
 
-- [二叉树的层序遍历](src/tree/binary-tree-right-side-view.js)
+- [二叉树的层序遍历](src/tree/binary-tree-level-order-traversal.js)
 
   - LeetCode 102. 二叉树的层序遍历 <https://leetcode-cn.com/problems/binary-tree-level-order-traversal/>
   - LintCode 69. 二叉树的层次遍历 <https://www.lintcode.com/problem/binary-tree-level-order-traversal/description>
 
+- [二叉树的层次遍历 II](src/tree/binary-tree-level-order-traversal-ii.js)
+
+  - LeetCode 107. 二叉树的层次遍历 II <https://leetcode-cn.com/problems/binary-tree-level-order-traversal-ii/>
+  - LintCode 70. 二叉树的层次遍历 II <https://www.lintcode.com/problem/binary-tree-level-order-traversal-ii/description>
+
 ## 链表
 
 - [合并K个排序链表](src/list/merge-k-sorted-lists.js)

src/tree/binary-tree-level-order-traversal-ii.js

@@ -0,0 +1,28 @@
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val) {
+ *     this.val = val;
+ *     this.left = this.right = null;
+ * }
+ */
+/**
+ * @param {TreeNode} root
+ * @return {number[][]}
+ */
+export const levelOrder = function (root) {
+  const res = []
+  const queue = [root]
+  while (queue.length) { // BFS
+    const tmp = []
+    const leave = queue.length // 记录这一层有几个
+    for (let i = 0; i < leave; i++) { // 一次性把固定个数的队列执行完
+      const node = queue.shift()
+      if (node && node.left) queue.push(node.left)
+      if (node && node.right) queue.push(node.right)
+      if (node) tmp.push(node.val)
+    }
+    if (tmp.length) res.unshift(tmp)
+  }
+
+  return res
+}

test/tree/binary-tree-level-order-traversal-ii.test.js

@@ -0,0 +1,14 @@
+import { levelOrder } from '../../src/tree/binary-tree-level-order-traversal-ii'
+import Tree from './Tree.js'
+
+test('二叉树的层序遍历 II', () => {
+  const source = [3, 9, 20, null, null, 15, 7]
+  expect(levelOrder(Tree.arrToTree(source))).toEqual([
+    [15, 7],
+    [9, 20],
+    [3]
+  ])
+
+  const source1 = []
+  expect(levelOrder(Tree.arrToTree(source1))).toEqual([])
+})