[抄题]:
Given a binary tree, determine if it is height-balanced.
For this problem, a height-balanced binary tree is defined as:
a binary tree in which the depth of the two subtrees of every node never differ by more than 1.
[思维问题]:
以为要用traverse一直判断,结果发现是没有读题。
[一句话思路]:
定义新的类 来帮助判断。(符合工业规范)
[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):
[画图]:
[一刷]:
最后的最大深度是left right中最大的,再+1
[二刷]:
[三刷]:
[四刷]:
[五刷]:
[总结]:
[复杂度]:Time complexity: O(n) Space complexity: O(lgn)
[英文数据结构,为什么不用别的数据结构]:
定义新的类 来帮助判断。(符合工业规范)
[其他解法]:
分治法
[Follow Up]:
[LC给出的题目变变变]:
/** * Definition of TreeNode: * public class TreeNode { * public int val; * public TreeNode left, right; * public TreeNode(int val) { * this.val = val; * this.left = this.right = null; * } * } */ public class Solution { /* * @param root: The root of binary tree. * @return: True if this Binary tree is Balanced, or false. */ class ResultType { boolean isBalanced; int maxDepth; ResultType(boolean isBalanced,int maxDepth) { this.isBalanced = isBalanced; this.maxDepth = maxDepth; } }; public boolean isBalanced(TreeNode root) { return helper(root).isBalanced; } private ResultType helper (TreeNode root) { if (root == null) { return new ResultType(true, 0); } ResultType left = helper(root.left); ResultType right = helper(root.right); if (!left.isBalanced || !right.isBalanced) { return new ResultType(false, - 1); } else if (Math.abs(left.maxDepth - right.maxDepth) > 1) { return new ResultType(false, - 1); } else { return new ResultType(true, Math.max(left.maxDepth,right.maxDepth) + 1); } } }