leetcode-4.28[107. 二叉樹的層次遍歷 II、101. 對稱二叉樹、104. 二叉樹的最大深度]（python解法）

題目1

題解1

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def levelOrderBottom(self, root: TreeNode) -> List[List[int]]:
        Nodes = []
        subList = []
        if root is None:
            return Nodes
        else:
            cNodeList = [root]
            nNodeList = []
        while True:
            if cNodeList:
                node = cNodeList.pop(0)
                subList.append(node.val)
                if node.left and node.right:
                    nNodeList.append(node.left)
                    nNodeList.append(node.right)
                elif node.left:
                    nNodeList.append(node.left)
                elif node.right:
                    nNodeList.append(node.right)
                else:
                    pass
            else:
                Nodes.insert(0,subList[:])
                subList = []
                if nNodeList:
                    cNodeList = nNodeList[:]
                    nNodeList = []
                else:
                    break
        return Nodes            

題目2

題解2

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def isSymmetric(self, root: TreeNode) -> bool:
        if root is None:
            return True
        else:
            return self.Symmetric(root.left,root.right)

    def Symmetric(self, lNode: TreeNode, rNode: TreeNode) -> bool:
        if lNode is None and rNode is None:
            return True
        elif lNode and rNode and lNode.val == rNode.val:
            return self.Symmetric(lNode.right, rNode.left) and self.Symmetric(lNode.left, rNode.right)
        else:
            return False

題目3

題解3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def maxDepth(self, root: TreeNode) -> int:
        if root is None:
            return 0
        else:
            return 1 + max(self.maxDepth(root.left), self.maxDepth(root.right))