Equal tree partition

Tree Depth-First Search Binary Tree

42.3% acceptance

Mar 31, 2026

508

Given the root of a binary tree, return true if you can partition the tree into two trees with equal sums of values after removing exactly one edge on the original tree.

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 check_equal_tree(root: Option<Rc<RefCell<TreeNode>>>) -> bool {
    fn subtree_sum(node: &Option<Rc<RefCell<TreeNode>>>, sums: &mut Vec<i64>) -> i64 {
      if let Some(n) = node {
        let n = n.borrow();
        let s = n.val as i64 + subtree_sum(&n.left, sums) + subtree_sum(&n.right, sums);
        sums.push(s);
        s
      } else {
        0
      }
    }
    let mut sums = Vec::new();
    let total = subtree_sum(&root, &mut sums);
    // Remove the last element (root's sum = total), check if any subtree sum == total/2
    sums.pop();
    if total % 2 != 0 { return false; }
    let half = total / 2;
    sums.iter().any(|&s| s == half)
  }
}