Update 0106.从中序与后序遍历序列构造二叉树.md

105 and 106 python version of code added

Author: jojoo15

problems/0106.从中序与后序遍历序列构造二叉树.md

@@ -618,8 +618,42 @@ class Solution {
 ```
 
 Python：
-
-
+105.从前序与中序遍历序列构造二叉树
+```python
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+//递归法
+class Solution:
+    def buildTree(self, preorder: List[int], inorder: List[int]) -> TreeNode:
+            if not preorder: return None  //特殊情况
+            root = TreeNode(preorder[0])  //新建父节点
+            p=inorder.index(preorder[0])  //找到父节点在中序遍历的位置(因为没有重复的元素，才可以这样找)
+            root.left = self.buildTree(preorder[1:p+1],inorder[:p])  //注意左节点时分割中序数组和前续数组的开闭环
+            root.right = self.buildTree(preorder[p+1:],inorder[p+1:])  //分割中序数组和前续数组
+            return root
+```
+106.从中序与后序遍历序列构造二叉树
+```python
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+//递归法
+class Solution:
+    def buildTree(self, inorder: List[int], postorder: List[int]) -> TreeNode:
+            if not postorder: return None  //特殊情况
+            root = TreeNode(postorder[-1]) //新建父节点
+            p=inorder.index(postorder[-1]) //找到父节点在中序遍历的位置*因为没有重复的元素，才可以这样找
+            root.left = self.buildTree(inorder[:p],postorder[:p]) //分割中序数组和后续数组
+            root.right = self.buildTree(inorder[p+1:],postorder[p:-1]) //注意右节点时分割中序数组和后续数组的开闭环
+            return root
+```
 Go：
 
 
@@ -643,4 +677,4 @@ var buildTree = function(inorder, postorder) {
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+<div align="center"><img src=../pics/公众号.png width=450 alt=> </img></div>