# 671. Second Minimum Node In a Binary Tree

## Description

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.

## Solutions

• /**
* 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);
}
}
}
}

• /**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode() : val(0), left(nullptr), right(nullptr) {}
*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int ans = -1;

int findSecondMinimumValue(TreeNode* root) {
dfs(root, root->val);
return ans;
}

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

• # 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


• /**
* 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
}

• /**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
*     this.val = (val===undefined ? 0 : val)
*     this.left = (left===undefined ? null : left)
*     this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {number}
*/
var findSecondMinimumValue = function (root) {
let ans = -1;
const v = root.val;
function dfs(root) {
if (!root) {
return;
}
dfs(root.left);
dfs(root.right);
if (root.val > v) {
if (ans == -1 || ans > root.val) {
ans = root.val;
}
}
}
dfs(root);
return ans;
};