树专题1 - leetcode101. Symmetric Tree/104. Maximum Depth of Binary Tree - Easy

101. Symmetric Tree

题目描述

判断是否是镜像二叉树（对称）

例子

思想
（递归）
定义一辅助函数，输入为left和right，判断是否相等。
（迭代）
定义两个栈，层次遍历时判断，并分别从左向右和从右向左存储。

解法1
递归

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

class Solution(object):
    def isSymmetric(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if not root:
            return True
        return self.helper(root.left, root.right)
    
    def helper(self, left, right):
        if not left and not right:
            return True
        if not left or not right or left.val != right.val:
            return False
        return self.helper(left.left, right.right) and self.helper(left.right, right.left)

解法2
迭代

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

class Solution(object):
    def isSymmetric(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if not root or not root.left and not root.right:
            return True
        if not root.left or not root.right:
            return False
        
        queue1, queue2 = [root.left], [root.right]
        while queue1 and queue2:
            node1 = queue1.pop(0)
            node2 = queue2.pop(0)
            
            if node1.val != node2.val:
                return False
            
            if node1.left and node2.right:
                queue1.append(node1.left)
                queue2.append(node2.right)
            elif node1.left or node2.right:
                return False
            
            if node1.right and node2.left:
                queue1.append(node1.right)
                queue2.append(node2.left)
            elif node1.right or node2.left:
                return False
        return True

104. Maximum Depth of Binary Tree

题目描述

二叉树的最大深度

例子

思想
（递归）
最大深度等于左右子树的最大深度加1。
（非递归）
层次遍历

解法1
递归

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

class Solution(object):
    def maxDepth(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        if not root:
            return 0
        return 1 + max(self.maxDepth(root.left), self.maxDepth(root.right))

解法2
层次遍历

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

class Solution(object):
    def maxDepth(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        depth = 0
        queue = [root]
        while queue and root:
            depth += 1
            for _ in range(len(queue)):
                node = queue.pop(0)
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)       
        return depth