Python世界：力扣题110，平衡二叉树判别，easy

任务背景

问题来自力扣题目：110 Balanced Binary Tree，大意如下：

Given a binary tree, determine if it is height-balanced。

翻译下，需求是：判断给定二叉数是否高度平衡。

思路分析

想练手下二叉树的遍历，结果在easy级上踩了坑，容我细细道来。注意本题中前置条件已默认是二叉树输入，不用考虑非二叉的输入场景。

于是，我把题意理解为，求该树中最小遍历深度和最大遍历深度，两者之差不应超过1.

python 复制代码

# Definition for a binary tree node.
class TreeNode(object):
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

depth = 0
depth_min = 2**31
depth_max = -1
class Solution(object):
    def isBalanced(self, root):
        """
        :type root: Optional[TreeNode]
        :rtype: bool
        """

        flag_res = True
        if root == None:
            return flag_res

        def recursive(node):
            global depth, depth_max, depth_min
            if node == None:
                depth_max = max(depth_max, depth)
                depth_min = min(depth_min, depth)
                return
            print(node.val)
            depth = depth + 1
            recursive(node.left)
            recursive(node.right)
            depth = depth - 1

        # 重要：每次需重新初始化全局变量，否则深度最值不准
        global depth, depth_max, depth_min
        depth_min = 2**31
        depth_max = -1
        recursive(root)
        if (depth_max - depth_min > 1):
            print(depth_max, depth_min)
            flag_res = False

        return flag_res

初步用例通过后，提交发现这个用例错误：

python 复制代码

# case true 错误理解，失败用例
root = TreeNode(1)
root.left = TreeNode(2)
root.right = TreeNode(3)
root.left.left = TreeNode(4)
root.left.right = TreeNode(5)
root.left.left.left = TreeNode(8)
root.right.right = TreeNode(3)
root.right.left = TreeNode(6)

才幡然醒悟，题意理解偏了，二叉树是否平衡，本质问的是：平衡树指每个节点的左右两个子树深度差异最大不超过2。

所以，我们应该求：对每个左右子树求取最大深度，比较左右子树差异。

代码实现

python 复制代码

# Definition for a binary tree node.
class TreeNode(object):
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

class Solution(object):
    def isBalanced(self, root):
        """
        :type root: Optional[TreeNode]
        :rtype: bool
        """
        # get the max depth of tree
        def recursive(node):
            if node == None:
                return 0
            depth_left = recursive(node.left)
            depth_right = recursive(node.right)
            if depth_left < 0 or depth_right < 0:
                return -1 # 已有子树不平衡
            if abs(depth_left - depth_right) > 1:
                return -1 # 当前不平衡
            return max(depth_left, depth_right) + 1 # 左右子树深度上提

        depth = recursive(root)
        return (depth >= 0) # 非负则平衡，负则不平衡

测试套件

单例测试版主调：

python 复制代码

# case true
root = TreeNode(1)

# case true
root = None

# case true
root = TreeNode(0)
root.left = TreeNode(1)
root.right = TreeNode(2)

# case false
root = TreeNode(1)
root.right = TreeNode(2)
root.right.left = TreeNode(3)

# case true
root = TreeNode(0)
root.left = TreeNode(1)
root.right = TreeNode(2)
root.right.right = TreeNode(3)

# case true 错误理解，失败用例
root = TreeNode(1)
root.left = TreeNode(2)
root.right = TreeNode(3)
root.left.left = TreeNode(4)
root.left.right = TreeNode(5)
root.left.left.left = TreeNode(8)
root.right.right = TreeNode(3)
root.right.left = TreeNode(6)


sol = Solution()
res = sol.isBalanced(root)
print(res)

测试套版主调：

python 复制代码

import unittest


def test_base(self, root, ret):
    sol = Solution()
    res = sol.isBalanced(root)
    self.assertEqual(res, ret)


# 编写测试套
class TestSol(unittest.TestCase):
    def test_special1(self):
        ret = True
        root = TreeNode(1)
        test_base(self, root, ret)

    def test_special2(self):
        ret = True
        root = None
        test_base(self, root, ret)

    def test_common1(self):
        ret = True
        root = TreeNode(0)
        root.left = TreeNode(1)
        root.right = TreeNode(2)
        test_base(self, root, ret)

    def test_common2(self):
        ret = False
        root = TreeNode(1)
        root.right = TreeNode(2)
        root.right.left = TreeNode(3)
        test_base(self, root, ret)

    def test_common3(self):
        ret = True
        root = TreeNode(0)
        root.left = TreeNode(1)
        root.right = TreeNode(2)
        root.right.right = TreeNode(3)
        test_base(self, root, ret)

    def test_common4(self):
        ret = True # 错误理解，失败用例
        root = TreeNode(1)
        root.left = TreeNode(2)
        root.right = TreeNode(3)
        root.left.left = TreeNode(4)
        root.left.right = TreeNode(5)
        root.left.left.left = TreeNode(8)
        root.right.right = TreeNode(3)
        root.right.left = TreeNode(6)


# 测试套版本主调
if __name__ == '__main__':
    print('start!')
    unittest.main() # 启动单元测试
    print('done!')

本文小结

呜乎，通过本文实践，简单的事情切不可大意，再次感受到：做正确的事，比正确的做事更重要！