second-minimum-node-in-a-binary-tree

Second Minimum Node In a Binary Tree

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findSecondMinimumValue

Given a non-empty special binary tree consisting of nodes with the non-negative value, where each node in this tree has exactly <code>two</code> or <code>zero</code> sub-node. If the node has two sub-nodes, then this node&#39;s value is the smaller value among its two sub-nodes. More formally, the property&nbsp;<code>root.val = min(root.left.val, root.right.val)</code>&nbsp;always holds.

Given such a binary tree, you need to output the second minimum value in the set made of all the nodes&#39; value in the whole tree.

If no such second minimum value exists, output -1 instead.

&nbsp;

&nbsp;
Example 1:
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/15/smbt1.jpg" style="width: 431px; height: 302px;" />
<pre>
Input: root = [2,2,5,null,null,5,7]
Output: 5
Explanation: The smallest value is 2, the second smallest value is 5.
</pre>

Example 2:
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/15/smbt2.jpg" style="width: 321px; height: 182px;" />
<pre>
Input: root = [2,2,2]
Output: -1
Explanation: The smallest value is 2, but there isn&#39;t any second smallest value.
</pre>

&nbsp;
Constraints:

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 25]</code>.</li>
	<li><code>1 &lt;= Node.val &lt;= 231 - 1</code></li>
	<li><code>root.val == min(root.left.val, root.right.val)</code>&nbsp;for each internal node of the tree.</li>
</ul>

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