<p>Given an integer array <code>nums</code>, return <em>the maximum result of </em><code>nums[i] XOR nums[j]</code>, where <code>0 &lt;= i &lt;= j &lt; n</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,10,5,25,2,8]
<strong>Output:</strong> 28
<strong>Explanation:</strong> The maximum result is 5 XOR 25 = 28.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [14,70,53,83,49,91,36,80,92,51,66,70]
<strong>Output:</strong> 127
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 2 * 10<sup>5</sup></code></li>
	<li><code>0 &lt;= nums[i] &lt;= 2<sup>31</sup> - 1</code></li>
</ul>

class Solution {
public:
    struct TreeNode {
        int val;
        TreeNode* children[2];
        TreeNode(int a) {
            val = a;
            for(int i = 0; i < 2; i++) children[i] = nullptr;
        }
    };

    TreeNode* root;

    int findMaximumXOR(vector<int>& nums) {
        root = new TreeNode(0);
        int c = 0;
        for(int a: nums) c = max(c, a);
        if (c == 0) return 0;
        int n = log2(c);
        TreeNode* t;
        for(int a: nums) {
            t = root;
            for(int i = n; i >= 0 ;i--) {
                int c = a / pow(2, i);
                a -= c * pow(2, i);
                if (!t->children[c]) t->children[c] = new TreeNode(c);
                t = t->children[c];
            }
        }
        queue<vector<TreeNode*>> q, q1;
        q.push({root, root});
        int ans = 0;
        for(int i = n; i >= 0; i--) {
            int c = q.size();
            q1 = queue<vector<TreeNode*>>();
            for(int j = 0; j < c; j++) {
                TreeNode* t1 = q.front()[0];
                TreeNode* t2 = q.front()[1];
                q.pop();
                if (t1->children[0] && t2->children[1]) q.push({t1->children[0], t2->children[1]});
                if (t1->children[1] && t2->children[0]) q.push({t1->children[1], t2->children[0]});
                if (t1->children[0] && t2->children[0]) q1.push({t1->children[0], t2->children[0]});
                if (t1->children[1] && t2->children[1]) q1.push({t1->children[1], t2->children[1]});
            }
            if (!q.empty()) {
                ans += pow(2, i);
            }
            else q = q1;
        }         
        return ans;
    }
};
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