LeetCodeWithPython 110. 平衡二叉树

题目地址

https://leetcode-cn.com/problems/balanced-binary-tree/

题目描述

代码初步

• 思路：分别计算左右两边子树的高度，最后判断左右子数的高度差的绝对值是否不超过1。
• 问题：超时emmm(( ▼-▼ )

# 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 isBalanced(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if not root:
            return True
        leftBranch = 0
        rightBranch = 0
        # 计算左边分支的高度
        leftStack = [root]
        while leftStack:
            for i in leftStack:
                if i != None:
                    leftBranch = leftBranch + 1
                    leftStack.pop()
                    leftStack.append(i.left)
                else:
                    break

        # 计算右边分支的高度
        rightStack = [root]

        while rightStack:
            for i in rightStack:
                if i != None:
                    rightBranch = rightBranch + 1
                    rightStack.pop()
                    rightStack.append(i.right)
                else:
                    break

                # 如果左右两边的高度差大于1，则不是平衡二叉树
        if (leftBranch - rightBranch) > 1 or (rightBranch - leftBranch) > 1:
            return False
        else:
            return True

代码欣赏

# 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 isBalanced(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        
        if not root:
            return True
        left_depth = self.get_depth(root.left)
        right_depth = self.get_depth(root.right)
        if abs(left_depth - right_depth) > 1:
            return False
        # 二叉树定义：一个二叉树每个节点 的左右两个子树的高度差的绝对值不超过1
        # 当左右子树的高度差小于1时，再往下每个节点判断下去
        else:
            return self.isBalanced(root.left) and self.isBalanced(root.right)

        # 得到树的高度      
    def get_depth(self,root):
        """
        :type root: TreeNode
        :rtype: int
        """
        if root is None:
            return 0
        else:
            return max(self.get_depth(root.left),self.get_depth(root.right))+1