From b655cb2730d1bc60f02d72cfd2918cb4d0789d3f Mon Sep 17 00:00:00 2001
From: Baturu <45113401+z80160280@users.noreply.github.com>
Date: Fri, 4 Jun 2021 02:10:07 -0700
Subject: [PATCH] =?UTF-8?q?Update=200101.=E5=AF=B9=E7=A7=B0=E4=BA=8C?=
 =?UTF-8?q?=E5=8F=89=E6=A0=91.md?=
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---
 problems/0101.对称二叉树.md | 72 ++++++++++++++++++++++++++++++++
 1 file changed, 72 insertions(+)

diff --git a/problems/0101.对称二叉树.md b/problems/0101.对称二叉树.md
index d8797d30..3dba1648 100644
--- a/problems/0101.对称二叉树.md
+++ b/problems/0101.对称二叉树.md
@@ -360,6 +360,78 @@ Java：
 
 Python：
 
+> 递归法
+```python
+class Solution:
+    def isSymmetric(self, root: TreeNode) -> bool:
+        if not root:
+            return True
+        return self.compare(root.left, root.right)
+        
+    def compare(self, left, right):
+        #首先排除空节点的情况
+        if left == None and right != None: return False
+        elif left != None and right == None: return False
+        elif left == None and right == None: return True
+        #排除了空节点，再排除数值不相同的情况
+        elif left.val != right.val: return False
+        
+        #此时就是：左右节点都不为空，且数值相同的情况
+        #此时才做递归，做下一层的判断
+        outside = self.compare(left.left, right.right) #左子树：左、 右子树：右
+        inside = self.compare(left.right, right.left) #左子树：右、 右子树：左
+        isSame = outside and inside #左子树：中、 右子树：中 （逻辑处理）
+        return isSame
+```
+
+> 迭代法： 使用队列
+```python
+import collections
+class Solution:
+    def isSymmetric(self, root: TreeNode) -> bool:
+        if not root:
+            return True
+        queue = collections.deque()
+        queue.append(root.left) #将左子树头结点加入队列
+        queue.append(root.right) #将右子树头结点加入队列
+        while queue: #接下来就要判断这这两个树是否相互翻转
+            leftNode = queue.popleft()
+            rightNode = queue.popleft()
+            if not leftNode and not rightNode: #左节点为空、右节点为空，此时说明是对称的
+                continue
+            
+            #左右一个节点不为空，或者都不为空但数值不相同，返回false
+            if not leftNode or not rightNode or leftNode.val != rightNode.val:
+                return False
+            queue.append(leftNode.left) #加入左节点左孩子
+            queue.append(rightNode.right) #加入右节点右孩子
+            queue.append(leftNode.right) #加入左节点右孩子
+            queue.append(rightNode.left) #加入右节点左孩子
+        return True
+```
+
+> 迭代法：使用栈
+```python
+class Solution:
+    def isSymmetric(self, root: TreeNode) -> bool:
+        if not root:
+            return True
+        st = [] #这里改成了栈
+        st.append(root.left)
+        st.append(root.right)
+        while st:
+            leftNode = st.pop()
+            rightNode = st.pop()
+            if not leftNode and not rightNode:
+                continue
+            if not leftNode or not rightNode or leftNode.val != rightNode.val:
+                return False
+            st.append(leftNode.left)
+            st.append(rightNode.right)
+            st.append(leftNode.right)
+            st.append(rightNode.left)
+        return True
+```
 
 Go：