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    "<h1>112. Path Sum</h1>\n",
    "<hr>\n",
    "\n",
    "<!--Copy Paste Leetcode statement between-->\n",
    "<p>Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.</p>\n",
    "\n",
    "<p><strong>Note:</strong>&nbsp;A leaf is a node with no children.</p>\n",
    "\n",
    "<p><strong>Example:</strong></p>\n",
    "\n",
    "<p>Given the below binary tree and <code>sum = 22</code>,</p>\n",
    "\n",
    "<pre>      <strong>5</strong>\n",
    "     <strong>/</strong> \\\n",
    "    <strong>4</strong>   8\n",
    "   <strong>/</strong>   / \\\n",
    "  <strong>11</strong>  13  4\n",
    " /  <strong>\\</strong>      \\\n",
    "7    <strong>2</strong>      1\n",
    "</pre>\n",
    "\n",
    "<p>return true, as there exist a root-to-leaf path <code>5-&gt;4-&gt;11-&gt;2</code> which sum is 22.</p>\n",
    "<!--Copy Paste Leetcode statement between-->\n",
    "\n",
    "<p>&nbsp;</p>\n",
    "<a href=\"https://leetcode.com/problems/path-sum/\">Source</a> \n",
    "<hr>"
   ]
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  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [],
   "source": [
    "# Definition for a binary tree node.\n",
    "class TreeNode:\n",
    "    def __init__(self, val=0, left=None, right=None):\n",
    "        self.val = val\n",
    "        self.left = left\n",
    "        self.right = right"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<h4>Code</h4>"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [],
   "source": [
    "def has_path_sum(root, amount):\n",
    "    \"\"\"Recursive approach.\"\"\"\n",
    "    if root is None:\n",
    "        return False\n",
    "    if root.left is None and root.right is None:\n",
    "        return root.val == amount\n",
    "    if root.left is None:                                # Technically\n",
    "        return has_path_sum(root.right, amount-root.val)    # this part is not needed\n",
    "    if root.right is None:                               # but \n",
    "        return has_path_sum(root.left, amount-root.val)     # speed up the algo a tiny bit\n",
    "    return (has_path_sum(root.left, amount-root.val) or has_path_sum(root.right, amount-root.val))"
   ]
  },
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   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<hr>\n",
    "<h4>Follow up:</h4>\n",
    "<p>Solve it both recursively and iteratively.</p>\n",
    "\n",
    "<h4>Code</h4>"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [],
   "source": [
    "def has_path_sum(root, amount):\n",
    "    \"\"\"Iterative approach using BFS traversal with a list.\"\"\"\n",
    "    if not root:\n",
    "        return False\n",
    "    queue = [(root, 0)]\n",
    "    for (node, prev_sum) in queue:\n",
    "        if not node:\n",
    "            continue\n",
    "        curr_sum = prev_sum + node.val\n",
    "        if not node.left and not node.right and curr_sum == amount:  # only check for leaf nodes\n",
    "            return True\n",
    "        queue.extend([(node.left, curr_sum), (node.right, curr_sum)])\n",
    "    return False"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [],
   "source": [
    "def has_path_sum(root, amount):\n",
    "    \"\"\"Iterative approach using BFS traversal with a deque (slower).\"\"\"\n",
    "    if not root:\n",
    "        return 0\n",
    "    queue = deque()\n",
    "    queue.append((root, 0))\n",
    "    while queue:\n",
    "        node, prev_sum = queue.popleft()\n",
    "        if not node:\n",
    "            continue\n",
    "        curr_sum = prev_sum + node.val\n",
    "        if not node.left and not node.right and curr_sum == amount:  # only checkfor leaf nodes\n",
    "            return True\n",
    "        queue.extend([(node.left, curr_sum), (node.right, curr_sum)])\n",
    "    return False"
   ]
  }
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