分享程序员开发的那些事...
更新时间：2022-08-22 08:36:46
4.1 实现一个函数，检查二叉树是否平衡。在这个问题中，平衡树的定义如下：任意一个结点，其两颗子树的高度差不超过1.
C++实现代码：
#include<iostream> #include<new> #include<cmath> using namespace std; //Definition for binary tree struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int x) : val(x), left(NULL), right(NULL) {} }; void createTree(TreeNode *&root) { int i; cin>>i; if(i!=0) { TreeNode *tmp=new TreeNode(i); root=tmp; createTree(root->left); createTree(root->right); } } int high(TreeNode *root) { if(root==NULL) return 0; if(root->left==NULL&&root->right) return 1; int lhigh=high(root->left); int rhigh=high(root->right); return (lhigh>=rhigh)?lhigh+1:rhigh+1; } bool isAVL(TreeNode *root) { if(root==NULL) return true; if(root->left==NULL&&root->right==NULL) return true; int lhigh=high(root->left); int rhigh=high(root->right); if(abs(lhigh-rhigh)<2) return isAVL(root->left)&&isAVL(root->right); return false; } int main() { TreeNode *root=NULL; createTree(root); cout<<high(root)<<endl; cout<<isAVL(root)<<endl; }