Find Distance in a Binary Tree
<p>Given the root of a binary tree and two integers <code>p</code> and <code>q</code>, return <em>the <strong>distance</strong> between the nodes of value </em><code>p</code><em> and value </em><code>q</code><em> in the tree</em>.</p>

<p>The <strong>distance</strong> between two nodes is the number of edges on the path from one to the other.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1700-1799/1740.Find%20Distance%20in%20a%20Binary%20Tree/images/binarytree.png" />
<pre>
<strong>Input:</strong> root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 0
<strong>Output:</strong> 3
<strong>Explanation:</strong> There are 3 edges between 5 and 0: 5-3-1-0.</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1700-1799/1740.Find%20Distance%20in%20a%20Binary%20Tree/images/binarytree.png" />
<pre>
<strong>Input:</strong> root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 7
<strong>Output:</strong> 2
<strong>Explanation:</strong> There are 2 edges between 5 and 7: 5-2-7.</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1700-1799/1740.Find%20Distance%20in%20a%20Binary%20Tree/images/binarytree.png" />
<pre>
<strong>Input:</strong> root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 5
<strong>Output:</strong> 0
<strong>Explanation:</strong> The distance between a node and itself is 0.</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>0 &lt;= Node.val &lt;= 10<sup>9</sup></code></li>
	<li>All <code>Node.val</code> are <strong>unique</strong>.</li>
	<li><code>p</code> and <code>q</code> are values in the tree.</li>
</ul>
Tree; Depth-First Search; Breadth-First Search; Hash Table; Binary Tree
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int findDistance(TreeNode root, int p, int q) {
        TreeNode g = lca(root, p, q);
        return dfs(g, p) + dfs(g, q);
    }

    private int dfs(TreeNode root, int v) {
        if (root == null) {
            return -1;
        }
        if (root.val == v) {
            return 0;
        }
        int left = dfs(root.left, v);
        int right = dfs(root.right, v);
        if (left == -1 && right == -1) {
            return -1;
        }
        return 1 + Math.max(left, right);
    }

    private TreeNode lca(TreeNode root, int p, int q) {
        if (root == null || root.val == p || root.val == q) {
            return root;
        }
        TreeNode left = lca(root.left, p, q);
        TreeNode right = lca(root.right, p, q);
        if (left == null) {
            return right;
        }
        if (right == null) {
            return left;
        }
        return root;
    }
}