diff --git a/problems/0101.对称二叉树.md b/problems/0101.对称二叉树.md
index d8797d30..b58cef2e 100644
--- a/problems/0101.对称二叉树.md
+++ b/problems/0101.对称二叉树.md
@@ -363,6 +363,54 @@ Python：
 
 Go：
 
+```go
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+// 递归
+func defs(left *TreeNode, right *TreeNode) bool {
+    if left == nil && right == nil {
+        return true;
+    };
+    if left == nil || right == nil {
+        return false;
+    };
+    if left.Val != right.Val {
+        return false;
+    }
+    return defs(left.Left, right.Right) && defs(right.Left, left.Right);
+}
+func isSymmetric(root *TreeNode) bool {
+    return defs(root.Left, root.Right);
+}
+
+// 迭代
+func isSymmetric(root *TreeNode) bool {
+    var queue []*TreeNode;
+    if root != nil {
+        queue = append(queue, root.Left, root.Right);
+    }
+    for len(queue) > 0 {
+        left := queue[0];
+        right := queue[1];
+        queue = queue[2:];
+        if left == nil && right == nil {
+            continue;
+        }
+        if left == nil || right == nil || left.Val != right.Val {
+            return false;
+        };
+        queue = append(queue, left.Left, right.Right, right.Left, left.Right);
+    }
+    return true;
+}
+```
+
 
 JavaScript
 ```javascript
diff --git a/problems/0104.二叉树的最大深度.md b/problems/0104.二叉树的最大深度.md
index 756afb68..5f0fe411 100644
--- a/problems/0104.二叉树的最大深度.md
+++ b/problems/0104.二叉树的最大深度.md
@@ -284,6 +284,55 @@ Python：
 
 Go：
 
+```go
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func max (a, b int) int {
+    if a > b {
+        return a;
+    }
+    return b;
+}
+// 递归
+func maxDepth(root *TreeNode) int {
+    if root == nil {
+        return 0;
+    }
+    return max(maxDepth(root.Left), maxDepth(root.Right)) + 1;
+}
+
+// 遍历
+func maxDepth(root *TreeNode) int {
+    levl := 0;
+    queue := make([]*TreeNode, 0);
+    if root != nil {
+        queue = append(queue, root);
+    }
+    for l := len(queue); l > 0; {
+        for ;l > 0;l-- {
+            node := queue[0];
+            if node.Left != nil {
+                queue = append(queue, node.Left);
+            }
+            if node.Right != nil {
+                queue = append(queue, node.Right);
+            }
+            queue = queue[1:];
+        }
+        levl++;
+        l = len(queue);
+    }
+    return levl;
+}
+
+```
+
 
 JavaScript
 ```javascript
diff --git a/problems/0111.二叉树的最小深度.md b/problems/0111.二叉树的最小深度.md
index 8ee15eac..48795722 100644
--- a/problems/0111.二叉树的最小深度.md
+++ b/problems/0111.二叉树的最小深度.md
@@ -301,6 +301,64 @@ class Solution:
 
 Go：
 
+```go
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func min(a, b int) int {
+    if a < b {
+        return a;
+    }
+    return b;
+}
+// 递归
+func minDepth(root *TreeNode) int {
+    if root == nil {
+        return 0;
+    }
+    if root.Left == nil && root.Right != nil {
+        return 1 + minDepth(root.Right);
+    }
+    if root.Right == nil && root.Left != nil {
+        return 1 + minDepth(root.Left);
+    }
+    return min(minDepth(root.Left), minDepth(root.Right)) + 1;
+}
+
+// 迭代
+
+func minDepth(root *TreeNode) int {
+    dep := 0;
+    queue := make([]*TreeNode, 0);
+    if root != nil {
+        queue = append(queue, root);
+    }
+    for l := len(queue); l > 0; {
+        dep++;
+        for ; l > 0; l-- {
+            node := queue[0];
+            if node.Left == nil && node.Right == nil {
+                return dep;
+            }
+            if node.Left != nil {
+                queue = append(queue, node.Left);
+            }
+            if node.Right != nil {
+                queue = append(queue, node.Right);
+            }
+            queue = queue[1:];
+        }
+        l = len(queue);
+    } 
+    return dep;
+}
+```
+
 
 JavaScript:
 
diff --git a/problems/0239.滑动窗口最大值.md b/problems/0239.滑动窗口最大值.md
index ed17157a..47383a47 100644
--- a/problems/0239.滑动窗口最大值.md
+++ b/problems/0239.滑动窗口最大值.md
@@ -263,6 +263,41 @@ class Solution {
 ```
 
 Python：
+```python
+class MyQueue: #单调队列（从大到小
+    def __init__(self):
+        self.queue = [] #使用list来实现单调队列
+    
+    #每次弹出的时候，比较当前要弹出的数值是否等于队列出口元素的数值，如果相等则弹出。
+    #同时pop之前判断队列当前是否为空。
+    def pop(self, value):
+        if self.queue and value == self.queue[0]:
+            self.queue.pop(0)#list.pop()时间复杂度为O(n),这里可以使用collections.deque()
+            
+    #如果push的数值大于入口元素的数值，那么就将队列后端的数值弹出，直到push的数值小于等于队列入口元素的数值为止。
+    #这样就保持了队列里的数值是单调从大到小的了。
+    def push(self, value):
+        while self.queue and value > self.queue[-1]:
+            self.queue.pop()
+        self.queue.append(value)
+        
+    #查询当前队列里的最大值 直接返回队列前端也就是front就可以了。
+    def front(self):
+        return self.queue[0]
+    
+class Solution:
+    def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:
+        que = MyQueue()
+        result = []
+        for i in range(k): #先将前k的元素放进队列
+            que.push(nums[i])
+        result.append(que.front()) #result 记录前k的元素的最大值
+        for i in range(k, len(nums)):
+            que.pop(nums[i - k]) #滑动窗口移除最前面元素
+            que.push(nums[i]) #滑动窗口前加入最后面的元素
+            result.append(que.front()) #记录对应的最大值
+        return result
+```
 
 
 Go：
diff --git a/problems/0347.前K个高频元素.md b/problems/0347.前K个高频元素.md
index 29963b48..71af618e 100644
--- a/problems/0347.前K个高频元素.md
+++ b/problems/0347.前K个高频元素.md
@@ -163,64 +163,33 @@ class Solution {
 
 Python：
 ```python
+#时间复杂度：O(nlogk)
+#空间复杂度：O(n)
+import heapq
 class Solution:
-    def sift(self, alist, low, high):
-        '''小根堆构建'''
-        i = low 
-        j = 2 * i + 1
-        tmp = alist[low]
-        while j <= high:
-            if j + 1 <= high and alist[j+1] <= alist[j]:
-                j += 1
-            if alist[j] < tmp:
-                alist[i] = alist[j]
-                i = j 
-                j = 2 * i + 1
-            else:
-                alist[i] = tmp
-                break
-        else:
-            alist[i] = tmp
-
-
-    def topK(self, nums, k):
-        # 建立小根堆
-        heap = nums[:k]
-        for i in range((k-2)//2, -1, -1):
-            self.sift(heap, i, k-1)
-        
-        # 把后续的k到len(nums)填充到小根堆里
-        for i in range(k, len(nums)):
-            if nums[i] >= heap[0]:
-                heap[0] = nums[i]
-                self.sift(heap, 0, k-1)
-
-        # 排序
-        for i in range(k-1, -1, -1):
-            heap[0], heap[i]= heap[i], heap[0]
-            self.sift(heap, 0, i-1)
-        return heap
-        
     def topKFrequent(self, nums: List[int], k: int) -> List[int]:
-        dict1 = dict()
-        for val in nums:
-            if val not in dict1:
-                dict1[val] = 1
-            else:
-                dict1[val] += 1
-        res = list()
-        ind = list()
-        for item in dict1:
-            res.append([dict1[item], item])
-        result = list()
-        heap = self.topK(res, k)
-        print(heap)
-        for val in heap:
-            result.append(val[1])
+        #要统计元素出现频率
+        map_ = {} #nums[i]:对应出现的次数
+        for i in range(len(nums)):
+            map_[nums[i]] = map_.get(nums[i], 0) + 1
+        
+        #对频率排序
+        #定义一个小顶堆，大小为k
+        pri_que = [] #小顶堆
+        
+        #用固定大小为k的小顶堆，扫面所有频率的数值
+        for key, freq in map_.items():
+            heapq.heappush(pri_que, (freq, key))
+            if len(pri_que) > k: #如果堆的大小大于了K，则队列弹出，保证堆的大小一直为k
+                heapq.heappop(pri_que)
+        
+        #找出前K个高频元素，因为小顶堆先弹出的是最小的，所以倒叙来输出到数组
+        result = [0] * k
+        for i in range(k-1, -1, -1):
+            result[i] = heapq.heappop(pri_que)[1]
         return result
 ```
 
-
 Go：
 
 
diff --git a/problems/0459.重复的子字符串.md b/problems/0459.重复的子字符串.md
index d9dfc62c..51a903ef 100644
--- a/problems/0459.重复的子字符串.md
+++ b/problems/0459.重复的子字符串.md
@@ -184,6 +184,55 @@ class Solution {
 
 Python：
 
+这里使用了前缀表统一减一的实现方式
+
+```python
+class Solution:
+    def repeatedSubstringPattern(self, s: str) -> bool:  
+        if len(s) == 0:
+            return False
+        nxt = [0] * len(s)
+        self.getNext(nxt, s)
+        if nxt[-1] != -1 and len(s) % (len(s) - (nxt[-1] + 1)) == 0:
+            return True
+        return False
+    
+    def getNext(self, nxt, s):
+        nxt[0] = -1
+        j = -1
+        for i in range(1, len(s)):
+            while j >= 0 and s[i] != s[j+1]:
+                j = nxt[j]
+            if s[i] == s[j+1]:
+                j += 1
+            nxt[i] = j
+        return nxt
+```
+
+前缀表（不减一）的代码实现
+
+```python
+class Solution:
+    def repeatedSubstringPattern(self, s: str) -> bool:  
+        if len(s) == 0:
+            return False
+        nxt = [0] * len(s)
+        self.getNext(nxt, s)
+        if nxt[-1] != 0 and len(s) % (len(s) - nxt[-1]) == 0:
+            return True
+        return False
+    
+    def getNext(self, nxt, s):
+        nxt[0] = 0
+        j = 0
+        for i in range(1, len(s)):
+            while j > 0 and s[i] != s[j]:
+                j = nxt[j - 1]
+            if s[i] == s[j]:
+                j += 1
+            nxt[i] = j
+        return nxt
+```
 
 Go：
 
diff --git a/problems/二叉树的统一迭代法.md b/problems/二叉树的统一迭代法.md
index f4091ad5..533bdfe7 100644
--- a/problems/二叉树的统一迭代法.md
+++ b/problems/二叉树的统一迭代法.md
@@ -239,7 +239,78 @@ Java：
 
   ```
 Python：
+> 迭代法前序遍历
 
+```python
+class Solution:
+    def preorderTraversal(self, root: TreeNode) -> List[int]:
+        result = []
+        st= []
+        if root:
+            st.append(root)
+        while st:
+            node = st.pop()
+            if node != None:
+                if node.right: #右
+                    st.append(node.right)
+                if node.left: #左
+                    st.append(node.left)
+                st.append(node) #中
+                st.append(None)
+            else:
+                node = st.pop()
+                result.append(node.val)
+        return result
+```
+
+> 迭代法中序遍历
+```python
+class Solution:
+    def inorderTraversal(self, root: TreeNode) -> List[int]:
+        result = []
+        st = []
+        if root:
+            st.append(root)
+        while st:
+            node = st.pop()
+            if node != None:
+                if node.right: #添加右节点（空节点不入栈）
+                    st.append(node.right)
+                
+                st.append(node) #添加中节点
+                st.append(None) #中节点访问过，但是还没有处理，加入空节点做为标记。
+                
+                if node.left: #添加左节点（空节点不入栈）
+                    st.append(node.left)
+            else: #只有遇到空节点的时候，才将下一个节点放进结果集
+                node = st.pop() #重新取出栈中元素
+                result.append(node.val) #加入到结果集
+        return result
+```
+
+> 迭代法后序遍历
+```python
+class Solution:
+    def postorderTraversal(self, root: TreeNode) -> List[int]:
+        result = []
+        st = []
+        if root:
+            st.append(root)
+        while st:
+            node = st.pop()
+            if node != None:
+                st.append(node) #中
+                st.append(None)
+                
+                if node.right: #右
+                    st.append(node.right)
+                if node.left: #左
+                    st.append(node.left)
+            else:
+                node = st.pop()
+                result.append(node.val)
+        return result
+```
 
 Go：
 > 前序遍历统一迭代法