0671-Second Minimum Node In a Binary Tree.md

Second Minimum Node In a Binary Tree

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: 
    2
   / \
  2   5
     / \
    5   7

Output: 5
Explanation: The smallest value is 2, the second smallest value is 5.

Example 2:

Input: 
    2
   / \
  2   2

Output: -1
Explanation: The smallest value is 2, but there isn't any second smallest value.

Solution

O(n) Time, O(1) Space - Recursive:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public var val: Int
 *     public var left: TreeNode?
 *     public var right: TreeNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.left = nil
 *         self.right = nil
 *     }
 * }
 */
class Solution {
    func findSecondMinimumValue(_ root: TreeNode?) -> Int {
        guard let node = root else { return Int.max }
        let result = secondMin(node, node.val)
        return result == Int.max ? -1 : result
    }
    
    func secondMin(_ node: TreeNode?, _ min: Int) -> Int {
        guard let node = node else { return Int.max }
        switch node.val {
            case min:
            switch (secondMin(node.left, min), secondMin(node.right, min)) {
                case (min, min):
                return Int.max
                case (min, let val):
                return val
                case (let val, min):
                return val
                case let (lhs, rhs):
                return Swift.min(lhs, rhs)
            }
            default:
            return node.val
        }
    }
}

O(n) Time, O(n/2) Space - Iterative:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public var val: Int
 *     public var left: TreeNode?
 *     public var right: TreeNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.left = nil
 *         self.right = nil
 *     }
 * }
 */
class Solution {
    func findSecondMinimumValue(_ root: TreeNode?) -> Int {
        guard let root = root else { return -1 }
        var queue = [root], min = root.val, result = Int.max
        while !queue.isEmpty {
            let curr = queue.removeFirst()
            if curr.val != min {
                result = Swift.min(result, curr.val)
            }
            if let left = curr.left {
                queue.append(left)
            }
            if let right = curr.right {
                queue.append(right)
            }
        }
        return result == Int.max ? -1 : result
    }
}