O(n) time solution - Accepted

• The first minimum is always going to be the first node in the tree as the children will either be equal or bigger. And if a node is encountered whose value is greater or equal to the current second minimum, further traversal is not required. One can write a solution based on this.

``````class Solution {
public int findSecondMinimumValue(TreeNode root) {
if (root.right == null || root.left == null) {
return -1;
}
int firstMin = root.val;
int secondMin = helper(root, firstMin, root.left.val > root.right.val ? root.left.val : root.right.val);

return firstMin == secondMin ? -1 : secondMin;
}

public int helper(TreeNode root, int firstMin, int secondMin) {
if (root == null || root.val >= secondMin) {
return secondMin;
}

int leftMin = Integer.MAX_VALUE, rightMin = Integer.MAX_VALUE;
int min = secondMin > root.val ? (root.val > firstMin ? root.val : secondMin) : secondMin;
leftMin = helper(root.left, firstMin, min);
rightMin = helper(root.right, firstMin, min);

return leftMin > rightMin ? rightMin : leftMin;
}
}
``````

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