代码随想训练营第十四天（Python）｜ 层序遍历 10 、● 226.翻转二叉树 、101.对称二叉树 2

层序遍历
1、迭代法，使用队列

class Solution:
    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
        res = []
        if root is None:
            return res
        queue = [root]
        while queue:
            n = len(queue)
            tmp = []
            for i in range(n):
                node = queue.pop(0)
                tmp.append(node.val)
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)
            res.append(tmp)
        return res

2、递归

class Solution:
    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
        levels = []
        if root is None:
            return levels
        self.helper(root, 0, levels)
        return levels

    def helper(self, node, level, levels):
        if not node:
            return
        if level == len(levels):
            levels.append([])
        levels[level].append(node.val)
        self.helper(node.left, level+1, levels)
        self.helper(node.right, level+1, levels)

226.翻转二叉树
1、递归前序遍历

class Solution:
    def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        if root is None:
            return root
        root.left, root.right = root.right, root.left
        self.invertTree(root.left)
        self.invertTree(root.right)
        return root

2、前序遍历的统一写法

class Solution:
    def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        stack = []
        if root:
            stack.append(root)
        while stack:
            node = stack.pop()
            if node:
                if node.right:
                    stack.append(node.right) # 右
                if node.left:
                    stack.append(node.left) # 左
                # 需要处理的节点
                stack.append(node)  # 中
                stack.append(None)  # 作标记
            else:
                # 处理节点
                node = stack.pop()
                node.left, node.right = node.right, node.left
        return root

101.对称二叉树 2
1、递归法

class Solution:
    def isSymmetric(self, root: Optional[TreeNode]) -> bool:
        if root is None:
            return True
        return self.compare(root.left, root.right)

    def compare(self, left, right):
        if not left and not right: # 一层一层排除
            return True
        if not (left and right):
            return False
        if left.val != right.val:
            return False
        return self.compare(left.left, right.right) and self.compare(left.right, right.left)

2、使用栈

class Solution:
    def isSymmetric(self, root: Optional[TreeNode]) -> bool:
        st = []
        if root is None:
            return True
        st.append(root.left)
        st.append(root.right)
        while st:
            rightNode = st.pop()
            leftNode = st.pop()
            if not leftNode and not rightNode:
                continue
            if not rightNode or not leftNode 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