<p>Given the <code>root</code> of a binary tree, the depth of each node is <strong>the shortest distance to the root</strong>.</p>

<p>Return <em>the smallest subtree</em> such that it contains <strong>all the deepest nodes</strong> in the original tree.</p>

<p>A node is called <strong>the deepest</strong> if it has the largest depth possible among any node in the entire tree.</p>

<p>The <strong>subtree</strong> of a node is a tree consisting of that node, plus the set of all descendants of that node.</p>

<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 nodes of the tree.
Notice that nodes 5, 3 and 2 contain the deepest nodes in the tree but node 2 is the smallest subtree among them, so we return it.
</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.
</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 node in the tree is 2, the valid subtrees are the subtrees of nodes 2, 1 and 0 but the subtree of node 2 is the smallest.
</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, 500]</code>.</li>
	<li><code>0 &lt;= Node.val &lt;= 500</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 1123: <a href="https://leetcode.com/problems/lowest-common-ancestor-of-deepest-leaves/" target="_blank">https://leetcode.com/problems/lowest-common-ancestor-of-deepest-leaves/</a></p>

/**
 * 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:
    TreeNode* subtreeWithAllDeepest(TreeNode* root) {
        vector<TreeNode*> temp;
        vector<vector<TreeNode*>> level;
        unordered_map<TreeNode*,TreeNode*> hash;
        queue<pair<TreeNode*,TreeNode*>> q;
        q.push({root,NULL});
        q.push({NULL,NULL});

        hash[root] = NULL;
        while (!q.empty())
        {
            auto [node,parent] = q.front();q.pop();
            if (node == NULL)
            {
                level.push_back(temp);
                temp.clear();

                if (q.size()>0)
                {
                    q.push({NULL,NULL});
                }
            }
            else
            {
                temp.push_back(node);
                hash[node] = parent;
                if (node->left)
                {
                    q.push({node->left,node});
                }

                if (node->right)
                {
                    q.push({node->right,node});
                }
            }
        }

        temp = level.back();
        queue<TreeNode*> qe;
        unordered_map<TreeNode*,int> vis;
        for (auto leaves : temp)
        {
            vis[leaves] = 1;
            qe.push(leaves);
        }

        while (!qe.empty())
        {
            int sz = qe.size();
            if (sz == 1)
            {
                return qe.front();
            }

            while (sz--)
            {
                auto node = qe.front();qe.pop();

                auto parent = hash[node];
                if (vis.find(parent) == vis.end())
                {
                    vis[parent] = 1;
                    qe.push(parent);
                }
            }
        }

        return NULL;
    }
};
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