<p>You are given <code>n</code> <strong>BST (binary search tree) root nodes</strong> for <code>n</code> separate BSTs stored in an array <code>trees</code> (<strong>0-indexed</strong>). Each BST in <code>trees</code> has <strong>at most 3 nodes</strong>, and no two roots have the same value. In one operation, you can:</p>

<ul>
	<li>Select two <strong>distinct</strong> indices <code>i</code> and <code>j</code> such that the value stored at one of the <strong>leaves </strong>of <code>trees[i]</code> is equal to the <strong>root value</strong> of <code>trees[j]</code>.</li>
	<li>Replace the leaf node in <code>trees[i]</code> with <code>trees[j]</code>.</li>
	<li>Remove <code>trees[j]</code> from <code>trees</code>.</li>
</ul>

<p>Return<em> the <strong>root</strong> of the resulting BST if it is possible to form a valid BST after performing </em><code>n - 1</code><em> operations, or</em><em> </em><code>null</code> <i>if it is impossible to create a valid BST</i>.</p>

<p>A BST (binary search tree) is a binary tree where each node satisfies the following property:</p>

<ul>
	<li>Every node in the node&#39;s left subtree has a value&nbsp;<strong>strictly less</strong>&nbsp;than the node&#39;s value.</li>
	<li>Every node in the node&#39;s right subtree has a value&nbsp;<strong>strictly greater</strong>&nbsp;than the node&#39;s value.</li>
</ul>

<p>A leaf is a node that has no children.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/08/d1.png" style="width: 450px; height: 163px;" />
<pre>
<strong>Input:</strong> trees = [[2,1],[3,2,5],[5,4]]
<strong>Output:</strong> [3,2,5,1,null,4]
<strong>Explanation:</strong>
In the first operation, pick i=1 and j=0, and merge trees[0] into trees[1].
Delete trees[0], so trees = [[3,2,5,1],[5,4]].
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/24/diagram.png" style="width: 450px; height: 181px;" />
In the second operation, pick i=0 and j=1, and merge trees[1] into trees[0].
Delete trees[1], so trees = [[3,2,5,1,null,4]].
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/24/diagram-2.png" style="width: 220px; height: 165px;" />
The resulting tree, shown above, is a valid BST, so return its root.</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/08/d2.png" style="width: 450px; height: 171px;" />
<pre>
<strong>Input:</strong> trees = [[5,3,8],[3,2,6]]
<strong>Output:</strong> []
<strong>Explanation:</strong>
Pick i=0 and j=1 and merge trees[1] into trees[0].
Delete trees[1], so trees = [[5,3,8,2,6]].
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/24/diagram-3.png" style="width: 240px; height: 196px;" />
The resulting tree is shown above. This is the only valid operation that can be performed, but the resulting tree is not a valid BST, so return null.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/08/d3.png" style="width: 430px; height: 168px;" />
<pre>
<strong>Input:</strong> trees = [[5,4],[3]]
<strong>Output:</strong> []
<strong>Explanation:</strong> It is impossible to perform any operations.
</pre>

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

<ul>
	<li><code>n == trees.length</code></li>
	<li><code>1 &lt;= n &lt;= 5 * 10<sup>4</sup></code></li>
	<li>The number of nodes in each tree is in the range <code>[1, 3]</code>.</li>
	<li>Each node in the input may have children but no grandchildren.</li>
	<li>No two roots of <code>trees</code> have the same value.</li>
	<li>All the trees in the input are <strong>valid BSTs</strong>.</li>
	<li><code>1 &lt;= TreeNode.val &lt;= 5 * 10<sup>4</sup></code>.</li>
</ul>

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def canMerge(self, trees: List[TreeNode]) -> Optional[TreeNode]:
        def dfs(node):
            if not node:
                return
            if not node.left and not node.right and node.val in hashmap:
                newTree = hashmap[node.val]
                node.left, node.right = newTree.left, newTree.right
                hashmap.pop(node.val)
            dfs(node.left)
            dfs(node.right)

        def validate(node, lower, upper):
            if not node:
                return True
            if not lower < node.val < upper:
                return False
            else:
                return validate(node.left, lower, node.val) and validate(node.right, node.val, upper)

        if len(trees) == 1:
            return trees[0]
        nodeset = set()
        for tree in trees:
            if tree.left:
                nodeset.add(tree.left.val)
            if tree.right:
                nodeset.add(tree.right.val)
            if not tree.left and not tree.right:
                nodeset.add(tree.val)
        
        node = None
        for tree in trees:
            if tree.val not in nodeset:
                node = tree
        if not node:
            return None

        hashmap = {tree.val : tree for tree in trees if tree!=node}

        dfs(node)

        return node if len(hashmap) == 0 and validate(node, float('-inf'), float('inf')) else None
        
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