Merge pull request #323 from z80160280/master

添加 0226 0101 0104 python版本

Author: youngyangyang04

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：
 

problems/0104.二叉树的最大深度.md

@@ -281,6 +281,111 @@ class Solution {
 
 Python：
 
+104.二叉树的最大深度
+> 递归法：
+```python
+class Solution:
+    def maxDepth(self, root: TreeNode) -> int:
+        return self.getDepth(root)
+        
+    def getDepth(self, node):
+        if not node:
+            return 0
+        leftDepth = self.getDepth(node.left) #左
+        rightDepth = self.getDepth(node.right) #右
+        depth = 1 + max(leftDepth, rightDepth) #中
+        return depth
+```
+> 递归法；精简代码
+```python
+class Solution:
+    def maxDepth(self, root: TreeNode) -> int:
+        if not root:
+            return 0
+        return 1 + max(self.maxDepth(root.left), self.maxDepth(root.right))
+```
+
+> 迭代法：
+```python
+import collections
+class Solution:
+    def maxDepth(self, root: TreeNode) -> int:
+        if not root:
+            return 0
+        depth = 0 #记录深度
+        queue = collections.deque()
+        queue.append(root)
+        while queue:
+            size = len(queue)
+            depth += 1
+            for i in range(size):
+                node = queue.popleft()
+                if node.left:
+                    queue.append(node.left)
+                if node.right:
+                    queue.append(node.right)
+        return depth
+```
+
+559.N叉树的最大深度
+> 递归法：
+```python
+class Solution:
+    def maxDepth(self, root: 'Node') -> int:
+        if not root:
+            return 0
+        depth = 0
+        for i in range(len(root.children)):
+            depth = max(depth, self.maxDepth(root.children[i]))
+        return depth + 1
+```
+
+> 迭代法：
+```python
+import collections
+class Solution:
+    def maxDepth(self, root: 'Node') -> int:
+        queue = collections.deque()
+        if root:
+            queue.append(root)
+        depth = 0 #记录深度
+        while queue:
+            size = len(queue)
+            depth += 1
+            for i in range(size):
+                node = queue.popleft()
+                for j in range(len(node.children)):
+                    if node.children[j]:
+                        queue.append(node.children[j])
+        return depth
+```
+
+> 使用栈来模拟后序遍历依然可以
+```python
+class Solution:
+    def maxDepth(self, root: 'Node') -> int:
+        st = []
+        if root:
+            st.append(root)
+        depth = 0
+        result = 0
+        while st:
+            node = st.pop()
+            if node != None:
+                st.append(node) #中
+                st.append(None)
+                depth += 1
+                for i in range(len(node.children)): #处理孩子
+                    if node.children[i]:
+                        st.append(node.children[i])
+                    
+            else:
+                node = st.pop()
+                depth -= 1
+            result = max(result, depth)
+        return result
+```
+
 
 Go：
 

problems/0226.翻转二叉树.md

@@ -230,6 +230,56 @@ class Solution {
 
 Python：
 
+> 递归法：前序遍历
+```python
+class Solution:
+    def invertTree(self, root: TreeNode) -> TreeNode:
+        if not root:
+            return None
+        root.left, root.right = root.right, root.left #中
+        self.invertTree(root.left) #左
+        self.invertTree(root.right) #右
+        return root
+```
+
+> 迭代法：深度优先遍历（前序遍历）
+```python
+class Solution:
+    def invertTree(self, root: TreeNode) -> TreeNode:
+        if not root:
+            return root
+        st = []
+        st.append(root)
+        while st:
+            node = st.pop()
+            node.left, node.right = node.right, node.left #中
+            if node.right:
+                st.append(node.right) #右
+            if node.left:
+                st.append(node.left) #左
+        return root
+```
+
+> 迭代法：广度优先遍历（层序遍历）
+```python
+import collections
+class Solution:
+    def invertTree(self, root: TreeNode) -> TreeNode:
+        queue = collections.deque() #使用deque()
+        if root:
+            queue.append(root)
+        while queue:
+            size = len(queue)
+            for i in range(size):
+                node = queue.popleft()
+                node.left, node.right = node.right, node.left #节点处理
+                if node.left:
+                    queue.append(node.left)
+                if node.right:
+                    queue.append(node.right)
+        return root
+```
+
 Go：
 ```Go
 func invertTree(root *TreeNode) *TreeNode {