Maximum XOR of two elements in an array

/*

Given an integer array nums, return the maximum result of nums[i] XOR nums[j], where 0 <= i <= j < n.

Example 1:
Input: nums = [3,10,5,25,2,8]
Output: 28
Explanation: The maximum result is 5 XOR 25 = 28.

Example 2:
Input: nums = [14,70,53,83,49,91,36,80,92,51,66,70]
Output: 127


Constraints:

1 <= nums.length <= 2 * 10^5
0 <= nums[i] <= 2^31 - 1

*/


struct BIT_TrieNode{
    BIT_TrieNode* children[2];
    BIT_TrieNode(){
        this->children[0]=NULL;
        this->children[1]=NULL;
    }
};

class Solution {
private:
    BIT_TrieNode* root;
public:
    void insert(int num){
        BIT_TrieNode* pCrawl=root;
        for(int i=31;i>=0;i--){
            int bit_state=(num >> i) & 1;   // CHECK IF ith bit is 1 or 0
            if(pCrawl->children[bit_state]==NULL){
                pCrawl->children[bit_state]=new BIT_TrieNode();
            }
            pCrawl=pCrawl->children[bit_state];
        }
    }

    int query(int num){
        BIT_TrieNode* pCrawl=root;
        int res=0;
        for(int i=31;i>=0;i--){
            int bit_state=(num >> i) & 1;
            if(pCrawl->children[bit_state^1]!=NULL){ // check if opposite bit is available, if yes, go for it since different bit results in 1 in XOR
                pCrawl=pCrawl->children[bit_state^1];
                res=res | (1 << i);  // turning bit on
            }
            else{
                pCrawl=pCrawl->children[bit_state];
            }
        }
        return res;
    }

    int findMaximumXOR(vector<int>& nums) {
        root=new BIT_TrieNode();
        int res=0;
        for(auto x: nums){
            insert(x);
        }
        for(auto x: nums){
            res=max(res,query(x)); // the thought process is that we already made the trie using bits, now for example 1 we know binary of 5 is 00000000000000000000000000000101 in 32 bit format. So, while iterating on this binary rep of 5, if we get a 1 we should take that as we see in query. We always try to take the opposite bit of the current bit as this will result in 1 and give max XOR
        }
        return res;
    }
};