LeetCode #671 — EASY

Second Minimum Node In a Binary Tree

Build confidence with an intuition-first walkthrough focused on tree fundamentals.

The Problem

Problem Statement

Given a non-empty special binary tree consisting of nodes with the non-negative value, where each node in this tree has exactly two or zero sub-node. If the node has two sub-nodes, then this node's value is the smaller value among its two sub-nodes. More formally, the property root.val = min(root.left.val, root.right.val) always holds.

Given such a binary tree, you need to output the second minimum value in the set made of all the nodes' value in the whole tree.

If no such second minimum value exists, output -1 instead.

Example 1:

Input: root = [2,2,5,null,null,5,7]
Output: 5
Explanation: The smallest value is 2, the second smallest value is 5.

Example 2:

Input: root = [2,2,2]
Output: -1
Explanation: The smallest value is 2, but there isn't any second smallest value.

Constraints:

The number of nodes in the tree is in the range [1, 25].
1 <= Node.val <= 2³¹ - 1
root.val == min(root.left.val, root.right.val) for each internal node of the tree.

Step 01

Brute Force Baseline

Problem summary: Given a non-empty special binary tree consisting of nodes with the non-negative value, where each node in this tree has exactly two or zero sub-node. If the node has two sub-nodes, then this node's value is the smaller value among its two sub-nodes. More formally, the property root.val = min(root.left.val, root.right.val) always holds. Given such a binary tree, you need to output the second minimum value in the set made of all the nodes' value in the whole tree. If no such second minimum value exists, output -1 instead.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Tree

Example 1

[2,2,5,null,null,5,7]

Example 2

[2,2,2]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #671: Second Minimum Node In a Binary Tree
/**
 * 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 {
    private int ans = -1;

    public int findSecondMinimumValue(TreeNode root) {
        dfs(root, root.val);
        return ans;
    }

    private void dfs(TreeNode root, int val) {
        if (root != null) {
            dfs(root.left, val);
            dfs(root.right, val);
            if (root.val > val) {
                ans = ans == -1 ? root.val : Math.min(ans, root.val);
            }
        }
    }
}

// Accepted solution for LeetCode #671: Second Minimum Node In a Binary Tree
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func findSecondMinimumValue(root *TreeNode) int {
	ans, v := -1, root.Val
	var dfs func(*TreeNode)
	dfs = func(root *TreeNode) {
		if root == nil {
			return
		}
		dfs(root.Left)
		dfs(root.Right)
		if root.Val > v {
			if ans == -1 || ans > root.Val {
				ans = root.Val
			}
		}
	}
	dfs(root)
	return ans
}

# Accepted solution for LeetCode #671: Second Minimum Node In a Binary Tree
# 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 findSecondMinimumValue(self, root: Optional[TreeNode]) -> int:
        def dfs(root):
            if root:
                dfs(root.left)
                dfs(root.right)
                nonlocal ans, v
                if root.val > v:
                    ans = root.val if ans == -1 else min(ans, root.val)

        ans, v = -1, root.val
        dfs(root)
        return ans

// Accepted solution for LeetCode #671: Second Minimum Node In a Binary Tree
struct Solution;
use rustgym_util::*;

trait SecondMinimum {
    fn find_second_minimum_value(&self, min: &mut Option<i32>) -> Option<i32>;
    fn option_min(a: Option<i32>, b: Option<i32>) -> Option<i32>;
}

impl SecondMinimum for TreeLink {
    fn find_second_minimum_value(&self, min: &mut Option<i32>) -> Option<i32> {
        if let Some(node) = self {
            let node = node.borrow();
            let left = &node.left;
            let right = &node.right;
            if min.is_none() {
                *min = Some(node.val);
            }
            if min.unwrap() == node.val {
                let min_left = Self::find_second_minimum_value(left, min);
                let min_right = Self::find_second_minimum_value(right, min);
                Self::option_min(min_left, min_right)
            } else {
                Some(node.val)
            }
        } else {
            None
        }
    }
    fn option_min(a: Option<i32>, b: Option<i32>) -> Option<i32> {
        match (a, b) {
            (Some(a), Some(b)) => Some(i32::min(a, b)),
            (Some(a), None) => Some(a),
            (None, Some(b)) => Some(b),
            (None, None) => None,
        }
    }
}

impl Solution {
    fn find_second_minimum_value(root: TreeLink) -> i32 {
        let mut min = None;
        if let Some(second_min) = root.find_second_minimum_value(&mut min) {
            second_min
        } else {
            -1
        }
    }
}

#[test]
fn test() {
    let root = tree!(2, tree!(2), tree!(5, tree!(5), tree!(7)));
    assert_eq!(Solution::find_second_minimum_value(root), 5);
    let root = tree!(2, tree!(2), tree!(2));
    assert_eq!(Solution::find_second_minimum_value(root), -1);
}

// Accepted solution for LeetCode #671: Second Minimum Node In a Binary Tree
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #671: Second Minimum Node In a Binary Tree
// /**
//  * 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 {
//     private int ans = -1;
// 
//     public int findSecondMinimumValue(TreeNode root) {
//         dfs(root, root.val);
//         return ans;
//     }
// 
//     private void dfs(TreeNode root, int val) {
//         if (root != null) {
//             dfs(root.left, val);
//             dfs(root.right, val);
//             if (root.val > val) {
//                 ans = ans == -1 ? root.val : Math.min(ans, root.val);
//             }
//         }
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(h)

Approach Breakdown

LEVEL ORDER

O(n) time

O(n) space

BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.

DFS TRAVERSAL

O(n) time

O(h) space

Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.

Shortcut: Visit every node once → O(n) time. Recursion depth = tree height → O(h) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.