member access within misaligned address - stl

Question

question link : https://leetcode.com/problems/cousins-in-binary-tree
my code :
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
bool isCousins(TreeNode* root, int x, int y) {
//breath first search will be better in this approach
queue< pair< TreeNode * , TreeNode *> > a;
queue< pair< TreeNode * , TreeNode *> > b; // first children..then parent
a.push({root,NULL});//root-> parent = NULL
bool check = false;
auto it = a.front();
auto it2 = a.front();
while(a.size() !=0 || b.size() !=0){
if(a.size() == 0 && b.size() != 0)
a.swap(b);
//if(b.size() == 0)
// b.push({NULL,NULL});
it = a.front();
if(it.first->left != NULL)
b.push({it.first->left,it.first});
if(it.first->right != NULL)
b.push({it.first->right,it.first});
if( it.first != NULL && (it.first->val == x || it.first->val == y)){
//we need to find both in this level
//if not found return false;
while(a.size() != 0){
a.pop();
it2 = a.front();
//this is the line 40
//here....
if(it2.first != NULL && (((it2.first)->val == x) || ((it2.first)->val == y))){//all nodes are unique in value
if(it2.second == it.second)
return false;//same parent ..so not cousin
return true;//cousim
}
}
//only one of them is found at present depth
return false;
}
if(a.size() !=0)
a.pop();
}
//both not found
return false;
}
};
runtime error : Line 40: Char 60: runtime error: member access within misaligned address 0xbebebebebebebebe for type 'TreeNode', which requires 8 byte alignment (solution.cpp)
0xbebebebebebebebe: note: pointer points here
SUMMARY: UndefinedBehaviorSanitizer: undefined-behavior prog_joined.cpp:49:60