标签：purpose eve pos more else btree ret abs fun
Implement a function to check if a binary tree is balanced. For the purpose of this question, a balanced tree is defined to be a tree such that the heights of the two subtrees of any node never differ by more than one.
这道题让我们检查一棵树是不是平衡的，根据题目平衡二叉树的定义我们可以对于每个node递归的检测一下它的substrees是否符合这个条件。见如下的代码：
height函数递归的计算出一棵树的高度，is_balanced?函数递归的判断每个node的子树是不是平衡。
class TreeNode attr_accessor :val, :left, :right def initialize(val) @val = val @left = nil @right = nil end end def is_balanced?(root) return true if root.nil? if (height(root.left) - height(root.right)).abs <= 1 return is_balanced?(root.left) && is_balanced?(root.right) else return false end end def height(root) return 0 if root.nil? left_height = height(root.left) right_height = height(root.right) return [left_height, right_height].max + 1 end
[Cracking the Coding Interview] 4.4 Check Balanced
标签：purpose eve pos more else btree ret abs fun
原文地址：https://www.cnblogs.com/infinitycoder/p/9189513.html
