Leetcode DAY 15: 层序遍 and 翻转二叉树 and 对称二叉树

102. 二叉树的层序遍历

class Solution:
    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
        if not root:
            return []
        result = []
        #用队列
        que = deque([root])
        while que:
            size = len(que)
            res = []
            for _ in range(size):
                cur = que.popleft()
                res.append(cur.val)
                if cur.left:
                    que.append(cur.left)
                if cur.right:
                    que.append(cur.right)
            result.append(res)
        return result

• 117.填充每个节点的下一个右侧节点指针II

  class Solution:
      def connect(self, root: 'Node') -> 'Node':
          if not root: 
              return root
          que = deque([root])
          while que:
              size = len(que)
              for i in range(size):
                  if i == 0:
                      cur = que.popleft()
                      pre = cur
                  else:
                      cur = que.popleft()
                      pre.next = cur
                      pre = pre.next
                  if cur.left:
                      que.append(cur.left)
                  if cur.right:
                      que.append(cur.right)
              cur.next = None
          return root
  

class Solution:
    def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
 
        cur = root
        if not cur:
            return root
        #前序遍历 中左右
        #中间的处理逻辑
        cur.left, cur.right = cur.right, cur.left
        self.invertTree(cur.left)
        self.invertTree(cur.right)
 
        return root

class Solution:
    def isSymmetric(self, root: Optional[TreeNode]) -> bool:
        
        def compare(leftT: Optional[TreeNode], rightT: Optional[TreeNode]) -> bool:
            #首先把节点为空的情况考虑一下
            if leftT != None and rightT == None:
                return False
            elif leftT == None and rightT != None:
                return False
            elif leftT == None and rightT == None:
                return True
            elif leftT.val != rightT.val:
                return False
            return compare(leftT.left, rightT.right) and compare(leftT.right, rightT.left)
 
 
        if not root:
            return True
        return compare(root.left, root.right)

需要写一个递归函数对根节点的左右子节点进行判断，首先考虑节点为空的情况，最后考虑节点不为空的情况。如果数值不相等，返回False。最后比较左节点的左子节点 and 右节点的右子节点是否相等