判断是否是轴对称二叉树

来源：互联网发布：网络终端机复原编辑：程序博客网时间：2024/04/29 12:46

算法思路：

其实思路不难，递归判断左节点的左孩子是否等于右节点的右孩子，并且左节点的右孩子等于右节点的左孩子

#include <iostream>using namespace std;struct BinNode {char data;BinNode *lchild;BinNode *rchild;};void Create(BinNode* &root){char ch;cin >> ch;if(ch == '#'){root = NULL;}else{root = new BinNode;root->data = ch;Create(root->lchild);Create(root->rchild);}}int JudgeTree(BinNode *p, BinNode *q){if(p==NULL && q==NULL){return 1;}else if(p->data==q->data && p!=NULL && q!=NULL)  //左子树的左孩子等于右子树的右孩子，左子树的右孩子等于右子树的左孩子{JudgeTree(p->lchild, q->rchild);JudgeTree(p->rchild, q->lchild);}else{return 0;}}void PreOrder(BinNode *root){if(root == NULL){return;}else{cout <<root->data;PreOrder(root->lchild);PreOrder(root->rchild);}}void Release(BinNode *root){if(root == NULL){return;}else{Release(root->lchild);Release(root->rchild);delete root;}}int main(){int flag = 0;BinNode *root = NULL;Create(root);PreOrder(root);if(root != NULL){   flag = JudgeTree(root->lchild, root->rchild);}cout <<'\n'<< flag << '\n';Release(root);return 0;}

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