Validate Binary Search Tree
<p>Given the <code>root</code> of a binary tree, <em>determine if it is a valid binary search tree (BST)</em>.</p>

<p>A <strong>valid BST</strong> is defined as follows:</p>

<ul>
	<li>The left <span data-keyword="subtree">subtree</span> of a node contains only nodes with keys&nbsp;<strong>strictly less than</strong> the node&#39;s key.</li>
	<li>The right subtree of a node contains only nodes with keys <strong>strictly greater than</strong> the node&#39;s key.</li>
	<li>Both the left and right subtrees must also be binary search trees.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0098.Validate%20Binary%20Search%20Tree/images/tree1.jpg" style="width: 302px; height: 182px;" />
<pre>
<strong>Input:</strong> root = [2,1,3]
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0098.Validate%20Binary%20Search%20Tree/images/tree2.jpg" style="width: 422px; height: 292px;" />
<pre>
<strong>Input:</strong> root = [5,1,4,null,null,3,6]
<strong>Output:</strong> false
<strong>Explanation:</strong> The root node&#39;s value is 5 but its right child&#39;s value is 4.
</pre>

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

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 10<sup>4</sup>]</code>.</li>
	<li><code>-2<sup>31</sup> &lt;= Node.val &lt;= 2<sup>31</sup> - 1</code></li>
</ul>
Tree; Depth-First Search; Binary Search Tree; Binary Tree
/**
 * 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:
    bool isValidBST(TreeNode* root) {
        TreeNode* prev = nullptr;
        function<bool(TreeNode*)> dfs = [&](TreeNode* root) {
            if (!root) {
                return true;
            }
            if (!dfs(root->left)) {
                return false;
            }
            if (prev && prev->val >= root->val) {
                return false;
            }
            prev = root;
            return dfs(root->right);
        };
        return dfs(root);
    }
};