<p>Given the <code>root</code> of a binary tree, return <em>the lowest common ancestor of its deepest leaves</em>.</p>

<p>Recall that:</p>

<ul>
	<li>The node of a binary tree is a leaf if and only if it has no children</li>
	<li>The depth of the root of the tree is <code>0</code>. if the depth of a node is <code>d</code>, the depth of each of its children is <code>d + 1</code>.</li>
	<li>The lowest common ancestor of a set <code>S</code> of nodes, is the node <code>A</code> with the largest depth such that every node in <code>S</code> is in the subtree with root <code>A</code>.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://s3-lc-upload.s3.amazonaws.com/uploads/2018/07/01/sketch1.png" style="width: 600px; height: 510px;" />
<pre>
<strong>Input:</strong> root = [3,5,1,6,2,0,8,null,null,7,4]
<strong>Output:</strong> [2,7,4]
<strong>Explanation:</strong> We return the node with value 2, colored in yellow in the diagram.
The nodes coloured in blue are the deepest leaf-nodes of the tree.
Note that nodes 6, 0, and 8 are also leaf nodes, but the depth of them is 2, but the depth of nodes 7 and 4 is 3.</pre>

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

<pre>
<strong>Input:</strong> root = [1]
<strong>Output:</strong> [1]
<strong>Explanation:</strong> The root is the deepest node in the tree, and it&#39;s the lca of itself.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> root = [0,1,3,null,2]
<strong>Output:</strong> [2]
<strong>Explanation:</strong> The deepest leaf node in the tree is 2, the lca of one node is itself.
</pre>

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

<ul>
	<li>The number of nodes in the tree will be in the range <code>[1, 1000]</code>.</li>
	<li><code>0 &lt;= Node.val &lt;= 1000</code></li>
	<li>The values of the nodes in the tree are <strong>unique</strong>.</li>
</ul>

<p>&nbsp;</p>
<p><strong>Note:</strong> This question is the same as 865: <a href="https://leetcode.com/problems/smallest-subtree-with-all-the-deepest-nodes/" target="_blank">https://leetcode.com/problems/smallest-subtree-with-all-the-deepest-nodes/</a></p>

class Solution {
private:
    TreeNode* solve2(TreeNode* root, const vector<int>& deepestNodes) {
        if (!root) return NULL;

        // If the current root is one of the deepest nodes, return it.
        for (int node : deepestNodes) {
            if (root->val == node) return root;
        }

        // Recursively find LCA in both left and right subtrees
        TreeNode* left_ans = solve2(root->left, deepestNodes);
        TreeNode* right_ans = solve2(root->right, deepestNodes);

        // If both left and right return a node, then current root is the LCA
        if (left_ans != NULL && right_ans != NULL) return root;

        // Otherwise, return whichever subtree contains the deepest node
        return left_ans ? left_ans : right_ans;
    }

    void solve1(TreeNode*& ans, TreeNode* root, int p) {
        if (!root) return;

        if(root->val==p){
            ans = root;
            return;
        }

        solve1(ans, root->left, p);
        solve1(ans, root->right, p);
    }

public:
    TreeNode* lcaDeepestLeaves(TreeNode* root) {
        map<int, vector<int>> mpp;
        queue<pair<TreeNode*, int>> q;
        q.push({root, 0});

        // Level-order traversal with depth tracking
        while (!q.empty()) {
            TreeNode* temp = q.front().first;
            int depth = q.front().second;
            q.pop();

            if (temp) {
                mpp[depth].push_back(temp->val);
                if (temp->left) q.push({temp->left, depth + 1});
                if (temp->right) q.push({temp->right, depth + 1});
            }
        }

        TreeNode* ans = NULL;
        auto deepest = mpp.rbegin();  // Get the deepest level
        
        if (deepest->second.size() == 1) {
            solve1(ans, root, deepest->second[0]);
            return ans;
        } else {
            return solve2(root, deepest->second);  // Pass all deepest nodes
        }
    }
};

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