Formatted question description: https://leetcode.ca/all/1740.html

# 1740. Find Distance in a Binary Tree

Medium

## Description

Given the root of a binary tree and two integers p and q, return the distance between the nodes of value p and value q in the tree.

The distance between two nodes is the number of edges on the path from one to the other.

Example 1: Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 0

Output: 3

Explanation: There are 3 edges between 5 and 0: 5-3-1-0.

Example 2: Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 7

Output: 2

Explanation: There are 2 edges between 5 and 7: 5-2-7.

Example 3: Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 5

Output: 0

Explanation: The distance between a node and itself is 0.

Constraints:

• The number of nodes in the tree is in the range [1, 10^4].
• 0 <= Node.val <= 10^9
• All Node.val are unique.
• p and q are values in the tree.

## Solution

First, find the lowest common ancestor of the two nodes with values p and q. Next, calculate the distances from the lowest common ancestor to the two nodes with values p and q, respectively. Finally, calculate the sum of the two distances and return.

##### Alternative-1

hashmap to store distance of every node-pair during finding LCA, then just map look up.

##### Alternative-2

Dist(n1, n2) = Dist(root, n1) + Dist(root, n2) - 2 * Dist(root, lca)

• ‘n1’ and ‘n2’ are the two given keys
• ‘root’ is root of given Binary Tree.
• ‘lca’ is lowest common ancestor of n1 and n2
• Dist(n1, n2) is the distance between n1 and n2.
/**
* 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;
*     }
* }
*/

// alternative: hashmap to store distance of every node-pair during finding LCA, then just map look up

class Solution {
public int findDistance(TreeNode root, int p, int q) {
if (p == q)
return 0;
TreeNode ancestor = lowestCommonAncestor(root, p, q);
return getDistance(ancestor, p) + getDistance(ancestor, q);
}

public TreeNode lowestCommonAncestor(TreeNode root, int p, int q) {
if (root == null)
return null;
if (root.val == p || root.val == q)
return root;
TreeNode left = lowestCommonAncestor(root.left, p, q);
TreeNode right = lowestCommonAncestor(root.right, p, q);
if (left != null && right != null)
return root;
return left == null ? right : left;
}

public int getDistance(TreeNode node, int val) {
if (node == null)
return -1;
if (node.val == val)
return 0;
int leftDistance = getDistance(node.left, val);
int rightDistance = getDistance(node.right, val);
int subDistance = Math.max(leftDistance, rightDistance);
return subDistance >= 0 ? subDistance + 1 : -1;
}
}