Balanced Binary Tree
Source
Problem Statement 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. Example Given binary tree A={3,9,20,#,#,15,7}, B={3,#,20,15,7} A) 3 B) 3 / \ \ 9 20 20 / \ / \ 15 7 15 7 The binary tree A is a height-balanced binary tree, but B is not.
题解1 - 递归
根据题意,平衡树的定义是两子树的深度差最大不超过1,显然使用递归进行分析较为方便。既然使用递归,那么接下来就需要分析递归调用的终止条件。和之前的 Maximum Depth of Binary Tree 类似,NULL == root必然是其中一个终止条件,返回0;根据题意还需的另一终止条件应为「左右子树高度差大于1」,但对应此终止条件的返回值是多少?——INT_MAX or INT_MIN?想想都不合适,为何不在传入参数中传入bool指针或者bool引用咧?并以此变量作为最终返回值,此法看似可行,先来看看最开始想到的这种方法。
C++ Recursion with extra bool variable
/** * Definition of TreeNode: * class TreeNode { * public: * int val; * TreeNode *left, *right; * TreeNode(int val) { * this->val = val; * this->left = this->right = NULL; * } * } */ class Solution { public: /** * @param root: The root of binary tree. * @return: True if this Binary tree is Balanced, or false. */ bool isBalanced(TreeNode *root) { if (NULL == root) { return true; } bool result = true; maxDepth(root, result); return result; } private: int maxDepth(TreeNode *root, bool &isBalanced) { if (NULL == root) { return 0; } int leftDepth = maxDepth(root->left, isBalanced); int rightDepth = maxDepth(root->right, isBalanced); if (abs(leftDepth - rightDepth) > 1) { isBalanced = false; // speed up the recursion process return INT_MAX; } return max(leftDepth, rightDepth) + 1; } };
源码解析
如果在某一次子树高度差大于1时,返回INT_MAX以减少不必要的计算过程,加速整个递归调用的过程。
初看起来上述代码好像还不错的样子,但是在看了九章的实现后,瞬间觉得自己弱爆了... 首先可以确定abs(leftDepth - rightDepth) > 1肯定是需要特殊处理的,如果返回-1呢?咋一看似乎在下一步返回max(leftDepth, rightDepth) + 1时会出错,再进一步想想,我们能否不让max...这一句执行呢?如果返回了-1,其接盘侠必然是leftDepth或者rightDepth中的一个,因此我们只需要在判断子树高度差大于1的同时也判断下左右子树深度是否为-1即可都返回-1,不得不说这种处理方法要精妙的多,赞!
C++
/** * forked from http://www.jiuzhang.com/solutions/balanced-binary-tree/ * Definition of TreeNode: * class TreeNode { * public: * int val; * TreeNode *left, *right; * TreeNode(int val) { * this->val = val; * this->left = this->right = NULL; * } * } */ class Solution { public: /** * @param root: The root of binary tree. * @return: True if this Binary tree is Balanced, or false. */ bool isBalanced(TreeNode *root) { return (-1 != maxDepth(root)); } private: int maxDepth(TreeNode *root) { if (NULL == root) { return 0; } int leftDepth = maxDepth(root->left); int rightDepth = maxDepth(root->right); if (leftDepth == -1 || rightDepth == -1 || abs(leftDepth - rightDepth) > 1) { return -1; } return max(leftDepth, rightDepth) + 1; } };
Java
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public boolean isBalanced(TreeNode root) { return maxDepth(root) != -1; } private int maxDepth(TreeNode root) { if (root == null) return 0; int leftDepth = maxDepth(root.left); int rightDepth = maxDepth(root.right); if (leftDepth == -1 || rightDepth == -1 || Math.abs(leftDepth - rightDepth) > 1) { return -1; } return 1 + Math.max(leftDepth, rightDepth); } }
源码分析
抓住两个核心:子树的高度以及高度之差,返回值应该包含这两种信息。
复杂度分析
遍历所有节点各一次,时间复杂度为 O(n), 使用了部分辅助变量,空间复杂度 O(1).