algorithms/102-binary-tree-level-order-traversal.ipynb

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    "### [二叉树的层序遍历](https://leetcode-cn.com/problems/binary-tree-level-order-traversal/)\n",
    "\n",
    "给你一个二叉树，请你返回其按 层序遍历 得到的节点值。 （即逐层地，从左到右访问所有节点）。\n",
    "\n",
    "#### 示例：\n",
    "\n",
    "二叉树：`[3,9,20,null,null,15,7]`,\n",
    "\n",
    "```\n",
    "    3\n",
    "   / \\\n",
    "  9  20\n",
    "    /  \\\n",
    "   15   7\n",
    "```\n",
    "\n",
    "返回其层序遍历结果：\n",
    "```\n",
    "[\n",
    "  [3],\n",
    "  [9,20],\n",
    "  [15,7]\n",
    "]\n",
    "```\n",
    "\n",
    "#### 相关题目\n",
    "- [102. 二叉树的层序遍历](https://leetcode-cn.com/problems/binary-tree-level-order-traversal/)\n",
    "- [103. 二叉树的锯齿形层序遍历](https://leetcode-cn.com/problems/binary-tree-zigzag-level-order-traversal/)\n",
    "- [107. 二叉树的层序遍历 II](https://leetcode-cn.com/problems/binary-tree-level-order-traversal-ii/)\n",
    "- [429. N叉树的层序遍历](https://leetcode-cn.com/problems/n-ary-tree-level-order-traversal/)\n",
    "- [剑指 Offer 32 - I. 从上到下打印二叉树](https://leetcode-cn.com/problems/cong-shang-dao-xia-da-yin-er-cha-shu-lcof/)\n",
    "- [剑指 Offer 32 - II. 从上到下打印二叉树 II](https://leetcode-cn.com/problems/cong-shang-dao-xia-da-yin-er-cha-shu-ii-lcof/)\n",
    "- [剑指 Offer 32 - III. 从上到下打印二叉树 III](https://leetcode-cn.com/problems/cong-shang-dao-xia-da-yin-er-cha-shu-iii-lcof/)\n",
    "- [剑指 Offer 55 - I. 二叉树的深度](https://leetcode-cn.com/problems/er-cha-shu-de-shen-du-lcof/)\n",
    "- [104. 二叉树的最大深度](https://leetcode-cn.com/problems/maximum-depth-of-binary-tree/)\n",
    "- [111. 二叉树的最小深度](https://leetcode-cn.com/problems/minimum-depth-of-binary-tree/)\n",
    "- [559. N叉树的最大深度](https://leetcode-cn.com/problems/maximum-depth-of-n-ary-tree/)\n",
    "- [589. N叉树的前序遍历](https://leetcode-cn.com/problems/n-ary-tree-preorder-traversal/)\n",
    "- [590. N叉树的后序遍历](https://leetcode-cn.com/problems/n-ary-tree-postorder-traversal/)"
   ]
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    "from typing import List\n",
    "\n",
    "# Definition for a binary tree node.\n",
    "class TreeNode:\n",
    "    def __init__(self, x):\n",
    "        self.val = x\n",
    "        self.left = None\n",
    "        self.right = None\n",
    "        \n",
    "class Solution:\n",
    "    def levelOrder(self, root: TreeNode) -> List[List[int]]:\n",
    "        if root is None:\n",
    "            return []\n",
    "        ans, queue = [], [root]\n",
    "        while len(queue) > 0:\n",
    "            tmp, size = [], len(queue)\n",
    "            for i in range(size):\n",
    "                node = queue.pop(0)\n",
    "                \n",
    "                if node.left is not None:\n",
    "                    queue.append(node.left)\n",
    "                if node.right is not None:\n",
    "                    queue.append(node.right)\n",
    "                tmp.append(node.val)\n",
    "            \n",
    "            ans.append(tmp)\n",
    "        return ans"
   ]
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    "### 剑指 Offer 32 - I. 从上到下打印二叉树\n",
    "\n",
    "从上到下打印出二叉树的每个节点，同一层的节点按照从左到右的顺序打印。\n",
    "\n",
    "#### 例如:\n",
    "\n",
    "给定二叉树: `[3,9,20,null,null,15,7]`,\n",
    "```\n",
    "    3\n",
    "   / \\\n",
    "  9  20\n",
    "    /  \\\n",
    "   15   7\n",
    "```\n",
    "返回：\n",
    "```\n",
    "[3,9,20,15,7]\n",
    "```\n",
    "\n",
    "#### 提示：\n",
    "- 节点总数 <= 1000"
   ]
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    "class Solution:\n",
    "    def levelOrder(self, root: TreeNode) -> List[List[int]]:\n",
    "        ans, queue = [], [root]\n",
    "        while len(queue) > 0:\n",
    "            node = queue.pop(0)\n",
    "            ans.append(node.val)\n",
    "            if node.left is not None:\n",
    "                queue.append(node.left)\n",
    "            if node.right is not None:\n",
    "                queue.append(node.right)\n",
    "        return ans"
   ]
  }
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