# 1740. Find Distance in a Binary Tree

## 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, 104].
• 0 <= Node.val <= 109
• All Node.val are unique.
• p and q are values in the tree.

## Solutions

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;
*     }
* }
*/
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;
}
}

• /**
* 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:
int findDistance(TreeNode* root, int p, int q) {
TreeNode* g = lca(root, p, q);
return dfs(g, p) + dfs(g, q);
}

TreeNode* lca(TreeNode* root, int p, int q) {
if (!root || root->val == p || root->val == q) return root;
TreeNode* left = lca(root->left, p, q);
TreeNode* right = lca(root->right, p, q);
if (!left) return right;
if (!right) return left;
return root;
}

int dfs(TreeNode* root, int v) {
if (!root) 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 + max(left, right);
}
};

• # 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 findDistance(self, root: Optional[TreeNode], p: int, q: int) -> int:
def lca(root, p, q):
if root is None or root.val in [p, q]:
return root
left = lca(root.left, p, q)
right = lca(root.right, p, q)
if left is None:
return right
if right is None:
return left
return root

def dfs(root, v):
if root is None:
return -1
if root.val == v:
return 0
left, right = dfs(root.left, v), dfs(root.right, v)
if left == right == -1:
return -1
return 1 + max(left, right)

g = lca(root, p, q)
return dfs(g, p) + dfs(g, q)


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func findDistance(root *TreeNode, p int, q int) int {
var lca func(root *TreeNode, p int, q int) *TreeNode
lca = func(root *TreeNode, p int, q int) *TreeNode {
if root == nil || root.Val == p || root.Val == q {
return root
}
left, right := lca(root.Left, p, q), lca(root.Right, p, q)
if left == nil {
return right
}
if right == nil {
return left
}
return root
}
var dfs func(root *TreeNode, v int) int
dfs = func(root *TreeNode, v int) int {
if root == nil {
return -1
}
if root.Val == v {
return 0
}
left, right := dfs(root.Left, v), dfs(root.Right, v)
if left == -1 && right == -1 {
return -1
}
return 1 + max(left, right)
}
g := lca(root, p, q)
return dfs(g, p) + dfs(g, q)
}