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LeetCode/LeetCode0105从前序与中序遍历序列构造二叉树-Construct-Binary-Tree-from-Preorder-and-Inorder-Traversal.ipynb

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+    "# LeetCode0105从前序与中序遍历序列构造二叉树 Construct Binary Tree from Preorder and Inorder Traversal\n",
+    "\n",
+    "[重建二叉树LeetCode](https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/)\n",
+    "\n",
+    ">题目描述:\n",
+    "根据一棵树的前序遍历与中序遍历构造二叉树。\n",
+    "注意:\n",
+    "你可以假设树中没有重复的元素。\n",
+    "例如，给出\n",
+    "\n",
+    "```\n",
+    "preorder = [3,9,20,15,7]\n",
+    "inorder = [9,3,15,20,7]\n",
+    "\n",
+    "返回二叉树：\n",
+    "    3\n",
+    "   / \\\n",
+    "  9  20\n",
+    "    /  \\\n",
+    "   15   7\n",
+    "   \n",
+    "preorder = [1,2,4,7,3,5,6,8]\n",
+    "inorder = [4,7,2,1,5,3,8,6]\n",
+    "\n",
+    "返回二叉树：\n",
+    "       1\n",
+    "      / \\\n",
+    "     2  3\n",
+    "    /  /  \\\n",
+    "   4  5   6\n",
+    "    \\    /\n",
+    "     7  8\n",
+    "```\n",
+    "\n",
+    ">思路：\n",
+    ">\n",
+    ">前序的第一个元素是根结点的值，在中序中找到该值，中序中该值的左边的元素是根结点的左子树，右边是右子树，然后递归的处理左边和右边\n",
+    ">提示：二叉树结点，以及对二叉树的各种操作\n",
+    "\n",
+    "首先要知道一个结论，前序/后序+中序序列可以唯一确定一棵二叉树，所以自然而然可以用来建树。\n",
+    "\n",
+    "有如下特征：\n",
+    "1. 前序中左起第一位`1`肯定是根结点，我们可以据此找到中序中根结点的位置`rootin`；\n",
+    "2. 中序中根结点左边就是左子树结点，右边就是右子树结点，即`[左子树结点，根结点，右子树结点]`，我们就可以得出左子树结点个数为`int left = rootin - leftin;`；\n",
+    "3. 前序中结点分布应该是：`[根结点，左子树结点，右子树结点]`；\n",
+    "4. 根据前一步确定的左子树个数，可以确定前序中左子树结点和右子树结点的范围；\n",
+    "5. 如果我们要前序遍历生成二叉树的话，下一层递归应该是：\n",
+    "    - 左子树：`root->left = pre_order(前序左子树范围，中序左子树范围，前序序列，中序序列);`；\n",
+    "    - 右子树：`root->right = pre_order(前序右子树范围，中序右子树范围，前序序列，中序序列);`。\n",
+    "6. 每一层递归都要返回当前根结点`root`；\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "前序：\n",
+      "[1, 2, 4, 7, 3, 5, 6, 8]\n",
+      "中序：\n",
+      "[4, 7, 2, 1, 5, 3, 8, 6]\n",
+      "后序：\n",
+      "[7, 4, 2, 5, 8, 6, 3, 1]\n"
+     ]
+    }
+   ],
+   "source": [
+    "# 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 buildTree(self, preorder, inorder):\n",
+    "        \"\"\"\n",
+    "        :type preorder: List[int]\n",
+    "        :type inorder: List[int]\n",
+    "        :rtype: TreeNode\n",
+    "        \"\"\"\n",
+    "        if not preorder or not inorder:\n",
+    "            return None\n",
+    "        x = preorder.pop(0)\n",
+    "        node = TreeNode(x)\n",
+    "        idx = inorder.index(x)\n",
+    "        \n",
+    "        node.left = self.buildTree(preorder[:idx],inorder[:idx])\n",
+    "        node.right = self.buildTree(preorder[idx:],inorder[idx+1:])\n",
+    "        #返回二叉树根节点\n",
+    "        return node\n",
+    "\n",
+    "#下面是打印前序、中序和后序\n",
+    "def Preorder(root):\n",
+    "    res = []\n",
+    "    if root: #节点为None时，跳过\n",
+    "        res.append(root.val)\n",
+    "        res = res + Preorder(root.left)\n",
+    "        res = res + Preorder(root.right)\n",
+    "    return res\n",
+    "\n",
+    "def Inorder(root):\n",
+    "    res = []\n",
+    "    if root: #节点为None时，跳过\n",
+    "        res = res + Inorder(root.left)\n",
+    "        res.append(root.val)\n",
+    "        res = res + Inorder(root.right)\n",
+    "    return res\n",
+    "\n",
+    "def Postorder(root):\n",
+    "    res = []\n",
+    "    if root: #节点为None时，跳过\n",
+    "        res += Postorder(root.left)\n",
+    "        res += Postorder(root.right)\n",
+    "        res.append(root.val)\n",
+    "    return res\n",
+    "    \n",
+    "# 测试：\n",
+    "solu = Solution()\n",
+    "# preorder = [3,9,20,15,7]\n",
+    "# inorder = [9,3,15,20,7]\n",
+    "\n",
+    "preorder = [1,2,4,7,3,5,6,8]\n",
+    "inorder = [4,7,2,1,5,3,8,6]\n",
+    "\n",
+    "node = solu.buildTree(preorder, inorder)\n",
+    "print(\"前序：\")\n",
+    "print(Preorder(node))\n",
+    "print(\"中序：\")\n",
+    "print(Inorder(node))\n",
+    "print(\"后序：\")\n",
+    "print(Postorder(node))"
+   ]
+  }
+ ],
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