LeetCode Entry

110. Balanced Binary Tree

08.02.2026 easy 2026 kotlin rust

Is tree balanced?

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Problem TLDR

Is tree balanced? #easy #dfs

Intuition

Solve the sub-problem for every node. Compare max depths for the left and right.

Approach

• we can use ‘marker’ depth as a boolean
• we can shortcircuit and don’t check the right subtree
• we can override values in a tree to golf the solutino
• BFS will not solve this: leafs can be at any heights, only max depths left&right for each node matters

Complexity

• Time complexity: \(O(n)\)

• Space complexity: \(O(log(n))\)

Code

// 0ms
    fun isBalanced(r: TreeNode?): Boolean = r?.run {
        val b = isBalanced(left) && isBalanced(right)
        val l = left?.`val`?:0; val r = right?.`val`?:0
        `val`= 1 + max(l,r); b && abs(l-r) < 2
    } ?: true

// 0ms
    pub fn is_balanced(r: Option<Rc<RefCell<TreeNode>>>) -> bool {
        fn d(r: &Option<Rc<RefCell<TreeNode>>>) -> Result<i8, ()> {
            let Some(n) = r else { return Ok(0) }; let n = n.borrow();
            let l = d(&n.left)?; let r = d(&n.right)?;
            if (l-r).abs() > 1 { Err(()) } else { Ok(1 + l.max(r)) }
        }
        d(&r).is_ok()
    }