103. Binary Tree Zigzag Level Order Traversal.ipynb

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    "Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).\n",
    "\n",
    "For example:\n",
    "Given binary tree [3,9,20,null,null,15,7],\n",
    "    3\n",
    "   / \\\n",
    "  9  20\n",
    "    /  \\\n",
    "   15   7\n",
    "return its zigzag level order traversal as:\n",
    "[\n",
    "  [3],\n",
    "  [20,9],\n",
    "  [15,7]\n",
    "]"
   ]
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   "source": [
    "class TreeNode(object):\n",
    "    def __init__(self, x):\n",
    "        self.val = x\n",
    "        self.left = None\n",
    "        self.right = None"
   ]
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    "class Solution(object):\n",
    "    def zigzagLevelOrder(self, root):\n",
    "        \"\"\"\n",
    "        :type root: TreeNode\n",
    "        :rtype: List[List[int]]\n",
    "        \"\"\"\n",
    "        if not root:\n",
    "            return []\n",
    "        res=[]\n",
    "        q=[root]\n",
    "        zig=1\n",
    "        while len(q)>0:\n",
    "            temp=[]\n",
    "            newq=[]\n",
    "            for node in q:\n",
    "                if node.left:\n",
    "                    newq.append(node.left)\n",
    "                if node.right:\n",
    "                    newq.append(node.right)\n",
    "            for node in q[::zig]:\n",
    "                temp.append(node.val)\n",
    "            q=newq\n",
    "            res.append(temp)\n",
    "            zig=zig*(-1)\n",
    "        return res\n",
    "                \n",
    "        \n",
    "        "
   ]
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    "t1=TreeNode(2)\n",
    "t2=TreeNode(0)\n",
    "t3=TreeNode(3)\n",
    "t1.left=t2\n",
    "t1.right=t3\n",
    "t4=TreeNode(4)\n",
    "t5=TreeNode(5)\n",
    "t6=TreeNode(6)\n",
    "t2.left=t4\n",
    "t2.right=t5\n",
    "t4.left=t6\n",
    "s=Solution()"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {
    "collapsed": true
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   "outputs": [],
   "source": [
    "s=Solution()"
   ]
  },
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   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
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   "outputs": [
    {
     "data": {
      "text/plain": [
       "[[2], [3, 0], [4, 5], [6]]"
      ]
     },
     "execution_count": 8,
     "metadata": {},
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   ],
   "source": [
    "s.zigzagLevelOrder(t1)"
   ]
  },
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