Find distance in a binary tree

Hash Table Tree Depth-First Search Breadth-First Search Binary Tree

74.3% acceptance

Mar 31, 2026

485

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.

Solution

Rust

Time O(n)

Space O(n)

LeetCode

solution.rs

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
// 
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
  pub fn find_distance(root: Option<Rc<RefCell<TreeNode>>>, p: i32, q: i32) -> i32 {
    if p == q { return 0; }
    fn lca(node: &Option<Rc<RefCell<TreeNode>>>, p: i32, q: i32) -> Option<Rc<RefCell<TreeNode>>> {
      if let Some(n) = node {
        let val = n.borrow().val;
        if val == p || val == q { return node.clone(); }
        let left = lca(&n.borrow().left, p, q);
        let right = lca(&n.borrow().right, p, q);
        if left.is_some() && right.is_some() { return node.clone(); }
        if left.is_some() { left } else { right }
      } else { None }
    }
    fn depth(node: &Option<Rc<RefCell<TreeNode>>>, target: i32, d: i32) -> i32 {
      if let Some(n) = node {
        if n.borrow().val == target { return d; }
        let l = depth(&n.borrow().left, target, d + 1);
        if l != -1 { return l; }
        depth(&n.borrow().right, target, d + 1)
      } else { -1 }
    }
    let ancestor = lca(&root, p, q);
    depth(&ancestor, p, 0) + depth(&ancestor, q, 0)
  }
}