# Java DFS solution

• This problem can be solved by divide and conquer. For a TreeNode `root`, there are 3 cases:
(1) `root == null`, return -1;
(2) `root` has no children, then the second minimum value does not exist, return -1 as well;
(3) `root` has 2 children: `left` and `right`. Let's first assume `left.val < right.val`. If `left` doesn't have any children, `right.val` is what we want since `right.val` is the max value in the right subtree and `right.val` is greater than `left.val`.
If `left` has two children, we need compare the second minimum value of the left subtree and the minimum value of the right subtree. The smaller value is the second minimum value of the tree.
If `left.val == right.val`, we need to compare the second minimum values of the both subtrees.

`````` public int findSecondMinimumValue(TreeNode root) {
if (root == null) return -1;
if (root.left == null && root.right == null) return -1;

int leftR = root.left.val;
int rightR = root.right.val;

if (root.left.val < root.right.val) {
leftR = findSecondMinimumValue(root.left);
} else if (root.left.val > root.right.val) {
rightR = findSecondMinimumValue(root.right);
} else {
leftR = findSecondMinimumValue(root.left);
rightR = findSecondMinimumValue(root.right);
}

if (leftR == -1 && rightR == -1) return -1;
else if (leftR == -1) return rightR;
else if (rightR == -1) return leftR;
return Math.min(leftR, rightR);
}
``````

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