28- prove safety offer- face questions symmetrical binary tree - recursive binary tree

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Title: 
	determining whether a given symmetrical binary tree. 
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Ideas: 
	1, the recursive method. 
	2, basic conditions: two trees are empty is true; a null tree, a tree is not null, is false; two different values of the root of the tree, to false. 
	3, further: determining a left subtree of the tree and the tree right subtree 2 is symmetrical, it is determined whether the right sub-tree 1 and tree 2 child left symmetrically. 
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#Include <the iostream> 
#include <string.h> 
#include <algorithm> 
#include <the cmath> 
#include <stdio.h> 
the using namespace STD; 

struct BinaryTreeNode { 
	int Val; 
	BinaryTreeNode * left; 
	BinaryTreeNode * right; 
	BinaryTreeNode (int X): 
			Val (X), left (NULL), right (NULL) { 
	} 
}; 
 
BOOL isSymmetrical (BinaryTreeNode * pRoot1, BinaryTreeNode pRoot2 *) {
    IF (pRoot1 && pRoot2 == == nullptr a nullptr a) { 
        return to true; 
    }
    if(pRoot1 == nullptr || pRoot2 == nullptr) return false;

    bool flag = false;
    if(pRoot1->val == pRoot2->val){
        flag = isSymmetrical(pRoot1->left,pRoot2->right) && isSymmetrical(pRoot1->right,pRoot2->left);
    }
    return flag;
}

bool isSymmetrical(BinaryTreeNode* pRoot){
    if(pRoot == nullptr) return false;
    return isSymmetrical(pRoot->left,pRoot->right);
}


int main(){
   BinaryTreeNode* node1 = new BinaryTreeNode(8);
   BinaryTreeNode* node2 = new BinaryTreeNode(6);
   BinaryTreeNode* node3 = new BinaryTreeNode(6);
   BinaryTreeNode* node4 = new BinaryTreeNode(5);
   BinaryTreeNode* node5 = new BinaryTreeNode(7);
   BinaryTreeNode* node6 = new BinaryTreeNode(7);
   BinaryTreeNode* node7 = new BinaryTreeNode(5);
   node1->left = node2;
   node1->right = node3;
   node2->left = node4;
   node2->right = node5;
   node3->left = node6;
   node3->right = node7;

   cout<<isSymmetrical(node1)<<endl;
}