优化 0257.二叉树的所有路径.md Python3解法

1. 优化Python3递归解法
2. 追加Python3迭代解法
3. 尽量遵守PEP8，增强可读性

problems/0257.二叉树的所有路径.md

@@ -371,25 +371,57 @@ class Solution {
 Python：
 ```Python
 class Solution:
+    """二叉树的所有路径 递归法"""
+
     def binaryTreePaths(self, root: TreeNode) -> List[str]:
-        path=[]
+        path, result = '', []
-        res=[]
+        self.traversal(root, path, result)
-        def backtrace(root, path):
+        return result
-            if not root:return 
+    
-            path.append(root.val)
+    def traversal(self, cur: TreeNode, path: List, result: List):
-            if (not root.left)and (not root.right):
+        path += str(cur.val)
-               res.append(path[:])
+        # 如果当前节点为叶子节点，添加路径到结果中
-            ways=[]
+        if not (cur.left or cur.right):
-            if root.left:ways.append(root.left)
+            result.append(path)
-            if root.right:ways.append(root.right)
+            return
-            for way in ways:
+            
-                backtrace(way,path)
+        if cur.left:
-                path.pop()
+            self.traversal(cur.left, path + '->', result)
-        backtrace(root,path)
+            
-        return ["->".join(list(map(str,i))) for i in res]
+        if cur.right:
+            self.traversal(cur.right, path + '->', result)
 
 ```
 
+```python
+from collections import deque
+
+
+class Solution:
+    """二叉树的所有路径 迭代法"""
+
+    def binaryTreePaths(self, root: TreeNode) -> List[str]:
+        # 题目中节点数至少为1
+        stack, path_st, result = deque([root]), deque(), []
+        path_st.append(str(root.val))
+
+        while stack:
+            cur = stack.pop()
+            path = path_st.pop()
+            # 如果当前节点为叶子节点，添加路径到结果中
+            if not (cur.left or cur.right):
+                result.append(path)
+            if cur.right:
+                stack.append(cur.right)
+                path_st.append(path + '->' + str(cur.right.val))
+            if cur.left:
+                stack.append(cur.left)
+                path_st.append(path + '->' + str(cur.left.val))
+
+        return result
+```
+
+
 Go：
 ```go
 func binaryTreePaths(root *TreeNode) []string {