diff --git a/problems/0129.求根到叶子节点数字之和.md b/problems/0129.求根到叶子节点数字之和.md
index b37270e2..17642793 100644
--- a/problems/0129.求根到叶子节点数字之和.md
+++ b/problems/0129.求根到叶子节点数字之和.md
@@ -165,7 +165,32 @@ public:
 Java： 
 
 Python：
+```python3
+class Solution:
+    def sumNumbers(self, root: TreeNode) -> int:
+        res = 0
+        path = []
+        def backtrace(root):
+            nonlocal res
+            if not root: return # 节点空则返回
+            path.append(root.val)
+            if not root.left and not root.right: # 遇到了叶子节点
+                res += get_sum(path)
+            if root.left: # 左子树不空
+                backtrace(root.left)
+            if root.right: # 右子树不空
+                backtrace(root.right)
+            path.pop()
 
+        def get_sum(arr):
+            s = 0
+            for i in range(len(arr)):
+                s = s * 10 + arr[i]
+            return s
+
+        backtrace(root)
+        return res
+```
 Go：
 
 JavaScript：
diff --git a/problems/0143.重排链表.md b/problems/0143.重排链表.md
index 62232051..76df63b7 100644
--- a/problems/0143.重排链表.md
+++ b/problems/0143.重排链表.md
@@ -222,7 +222,61 @@ public class ReorderList {
 ```
 
 Python：
+```python3
+# 方法二 双向队列
+class Solution:
+    def reorderList(self, head: ListNode) -> None:
+        """
+        Do not return anything, modify head in-place instead.
+        """
+        d = collections.deque()
+        tmp = head
+        while tmp.next: # 链表除了首元素全部加入双向队列
+            d.append(tmp.next)
+            tmp = tmp.next
+        tmp = head
+        while len(d): # 一后一前加入链表
+            tmp.next = d.pop()
+            tmp = tmp.next
+            if len(d):
+                tmp.next = d.popleft()
+                tmp = tmp.next
+        tmp.next = None # 尾部置空
+ 
+# 方法三 反转链表
+class Solution:
+    def reorderList(self, head: ListNode) -> None:
+        if head == None or head.next == None:
+            return True
+        slow, fast = head, head
+        while fast and fast.next:
+            slow = slow.next
+            fast = fast.next.next
+        right = slow.next # 分割右半边
+        slow.next = None # 切断
+        right = self.reverseList(right) #反转右半边
+        left = head
+        # 左半边一定比右半边长, 因此判断右半边即可
+        while right:
+            curLeft = left.next
+            left.next = right
+            left = curLeft
 
+            curRight = right.next
+            right.next = left
+            right = curRight
+
+
+    def reverseList(self, head: ListNode) -> ListNode:
+        cur = head   
+        pre = None
+        while(cur!=None):
+            temp = cur.next # 保存一下cur的下一个节点
+            cur.next = pre # 反转
+            pre = cur
+            cur = temp
+        return pre
+```
 Go：
 
 JavaScript：
diff --git a/problems/0234.回文链表.md b/problems/0234.回文链表.md
index 6a24b1d0..b3ad899c 100644
--- a/problems/0234.回文链表.md
+++ b/problems/0234.回文链表.md
@@ -148,7 +148,62 @@ public:
 
 ## Python
 
-```python
+```python3
+#数组模拟
+class Solution:
+    def isPalindrome(self, head: ListNode) -> bool:
+        length = 0
+        tmp = head
+        while tmp: #求链表长度
+            length += 1
+            tmp = tmp.next
+        
+        result = [0] * length
+        tmp = head
+        index = 0
+        while tmp: #链表元素加入数组
+            result[index] = tmp.val
+            index += 1
+            tmp = tmp.next
+        
+        i, j = 0, length - 1
+        while i < j: # 判断回文
+            if result[i] != result[j]:
+                return False
+            i += 1
+            j -= 1
+        return True
+        
+#反转后半部分链表
+class Solution:
+    def isPalindrome(self, head: ListNode) -> bool:
+        if head == None or head.next == None:
+            return True
+        slow, fast = head, head
+        while fast and fast.next:
+            pre = slow
+            slow = slow.next
+            fast = fast.next.next
+        
+        pre.next = None # 分割链表
+        cur1 = head # 前半部分
+        cur2 = self.reverseList(slow) # 反转后半部分，总链表长度如果是奇数，cur2比cur1多一个节点
+        while cur1:
+            if cur1.val != cur2.val:
+                return False
+            cur1 = cur1.next
+            cur2 = cur2.next
+        return True
+
+    def reverseList(self, head: ListNode) -> ListNode:
+        cur = head   
+        pre = None
+        while(cur!=None):
+            temp = cur.next # 保存一下cur的下一个节点
+            cur.next = pre # 反转
+            pre = cur
+            cur = temp
+        return pre
 ```
 
 ## Go
diff --git a/problems/0724.寻找数组的中心索引.md b/problems/0724.寻找数组的中心索引.md
index 3ed68d47..b4115893 100644
--- a/problems/0724.寻找数组的中心索引.md
+++ b/problems/0724.寻找数组的中心索引.md
@@ -89,7 +89,16 @@ class Solution {
 
 ## Python
 
-```python
+```python3
+class Solution:
+    def pivotIndex(self, nums: List[int]) -> int:
+        numSum = sum(nums) #数组总和
+        leftSum = 0
+        for i in range(len(nums)):
+            if numSum - leftSum -nums[i] == leftSum: #左右和相等 
+                return i
+            leftSum += nums[i]
+        return -1
 ```
 
 ## Go
diff --git a/problems/0922.按奇偶排序数组II.md b/problems/0922.按奇偶排序数组II.md
index 92db204d..97d7091e 100644
--- a/problems/0922.按奇偶排序数组II.md
+++ b/problems/0922.按奇偶排序数组II.md
@@ -149,7 +149,32 @@ class Solution {
 
 ## Python
 
-```python
+```python3
+#方法2
+class Solution:
+    def sortArrayByParityII(self, nums: List[int]) -> List[int]:
+        result = [0]*len(nums)
+        evenIndex = 0
+        oddIndex = 1
+        for i in range(len(nums)):
+            if nums[i] % 2: #奇数
+                result[oddIndex] = nums[i]
+                oddIndex += 2
+            else: #偶数
+                result[evenIndex] = nums[i]
+                evenIndex += 2
+        return result
+
+#方法3
+class Solution:
+    def sortArrayByParityII(self, nums: List[int]) -> List[int]:
+        oddIndex = 1
+        for i in range(0,len(nums),2): #步长为2
+            if nums[i] % 2: #偶数位遇到奇数
+                while  nums[oddIndex] % 2: #奇数位找偶数
+                    oddIndex += 2
+                nums[i], nums[oddIndex] = nums[oddIndex], nums[i]
+        return nums
 ```
 
 ## Go