From 21757f5ab7650c06fa4cb39dc13d91f4818926f3 Mon Sep 17 00:00:00 2001
From: Kyle Liu <kylesliu@outlook.com>
Date: Thu, 24 Dec 2020 16:09:43 +0000
Subject: [PATCH] GitBook: [master] 196 pages modified

---
 docs/jzof/of004.md |   2 +-
 docs/jzof/of005.md |   2 +-
 docs/jzof/of007.md | 122 ++++++++++++++++++++++++++-------------------
 3 files changed, 72 insertions(+), 54 deletions(-)

diff --git a/docs/jzof/of004.md b/docs/jzof/of004.md
index e1a9eecaa..1701a114a 100644
--- a/docs/jzof/of004.md
+++ b/docs/jzof/of004.md
@@ -2,7 +2,7 @@
 description: 剑指 Offer 04. 二维数组中的查找
 ---
 
-# OF4. 二维数组中的查找
+# OF4.二维数组中的查找
 
 ## 题目描述
 
diff --git a/docs/jzof/of005.md b/docs/jzof/of005.md
index 4e68bb45c..0098111f7 100644
--- a/docs/jzof/of005.md
+++ b/docs/jzof/of005.md
@@ -2,7 +2,7 @@
 description: 剑指 Offer 05. 替换空格
 ---
 
-# OF5. 替换空格
+# OF5.替换空格
 
 ## 题目描述
 
diff --git a/docs/jzof/of007.md b/docs/jzof/of007.md
index 70ccd9540..702e621ba 100644
--- a/docs/jzof/of007.md
+++ b/docs/jzof/of007.md
@@ -2,7 +2,7 @@
 description: 剑指 Offer 07. 重建二叉树
 ---
 
-# OF6.重建二叉树
+# OF7.重建二叉树
 
 ## 题目描述
 
@@ -10,83 +10,101 @@ description: 剑指 Offer 07. 重建二叉树
 
 输入某二叉树的前序遍历和中序遍历的结果，请重建该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。
 
-
 ### **示例 1：**
 
 ```go
 前序遍历 preorder = [3,9,20,15,7]
 中序遍历 inorder = [9,3,15,20,7]
 
-3
-/ \
-9  20
-/  \
-15   7
+     3
+    / \
+   9  20
+   /   \
+  15    7
 ```
 
 ## 题解
 
-### 思路1 ： 递归
+### 思路1 ： ...
 
 递归遍历
 
 **算法流程：**
 
 1. **复杂度分析：**
-2. **时间复杂度**$$O(N)$$**：**遍历N次，递归 N 次
-3. **空间复杂度**$$O(N)$$**：**递归 N 次，开辟 N 个栈空间
+2. **时间复杂度**$$O(N)$$**：**
+3. **空间复杂度**$$O(N)$$**：**
 
 #### 代码
 
 {% tabs %}
 {% tab title="Go" %}
 ```go
-func reversePrint(head *ListNode) []int {
-    ans := make([]int, 0)
-    if head == nil {
-        return ans
-    }
-    ans = reversePrint(head.Next)
-    ans = append(ans, head.Val)
-    return ans
+type TreeNode struct {
+	Val   int
+	Left  *TreeNode
+	Right *TreeNode
+}
+
+//	递归求解
+func buildTree(pre []int, in []int) *TreeNode {
+	if len(pre) == 0 || len(in) == 0 {
+		return nil
+	}
+	mid := search(in, pre[0])
+	return &TreeNode{
+		Val:   pre[0],
+		Left:  buildTree(pre[1:mid+1], in[:mid+1]),
+		Right: buildTree(pre[mid+1:], in[mid+1:]),
+	}
+}
+func search(nodes []int, val int) int {
+	for p, v := range nodes {
+		if v == val {
+			return p
+		}
+	}
+	return -1
 }
 ```
 {% endtab %}
-{% endtabs %}
-
-### 思路1 ： 多指针
-
-多个指针辅助，一次遍历
-
-**算法流程：**
 
-1. **复杂度分析：**
-2. **时间复杂度**$$O(N)$$**：**遍历N次，递归 N 次
-3. **空间复杂度**$$O(N)$$**：**递归 N 次，开辟 N 个栈空间
-
-#### 代码
-
-{% tabs %}
-{% tab title="Go" %}
-```go
-func reversePrint(head *ListNode) []int {
-    if head == nil {
-        return []int{}
-    }
-    pre, cur, next, ans := &ListNode{}, head, head.Next, []int{}
-    for cur != nil {
-        next = cur.Next
-        cur.Next = pre
-
-        pre = cur
-        cur = next
-    }
-    for pre.Next != nil {
-        ans = append(ans, pre.Val)
-        pre = pre.Next
-    }
-    return ans
-}
+{% tab title="Python" %}
+```python
+class Solution:
+    def buildTree(self, preorder, inorder):
+        """
+        :type preorder: List[int]
+        :type inorder: List[int]
+        :rtype: TreeNode
+        """
+        def helper(in_left = 0, in_right = len(inorder)):
+            nonlocal pre_idx
+            # if there is no elements to construct subtrees
+            if in_left == in_right:
+                return None
+            
+            # pick up pre_idx element as a root
+            root_val = preorder[pre_idx]
+            root = TreeNode(root_val)
+
+            # root splits inorder list
+            # into left and right subtrees
+            index = idx_map[root_val]
+
+            # recursion 
+            pre_idx += 1
+            # build left subtree
+            root.left = helper(in_left, index)
+            # build right subtree
+            root.right = helper(index + 1, in_right)
+            return root
+        
+        # start from first preorder element
+        pre_idx = 0
+        # build a hashmap value -> its index
+        idx_map = {val:idx for idx, val in enumerate(inorder)} 
+        return helper()
 ```
 {% endtab %}
 {% endtabs %}