LeetCode //C - 671. Second Minimum Node In a Binary Tree

671. 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: 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].
• $1<=Node.val<=2^{31}-1$
• root.val == min(root.left.val, root.right.val) for each internal node of the tree.

From: LeetCode
Link: 671. Second Minimum Node In a Binary Tree

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Solution:

Ideas:

• Constraint given: root->val == min(root->left->val, root->right->val)

• So the root always contains the smallest value in the tree.

• We do a DFS to search for the smallest value larger than root->val.

• If a node has a value greater than root->val, it’s a candidate for second minimum.

• Otherwise, keep searching its children.

Code:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
int dfs(struct TreeNode* root, int minVal) {
    if (!root) return -1;
    
    if (root->val > minVal) {
        return root->val;
    }

    int left = dfs(root->left, minVal);
    int right = dfs(root->right, minVal);

    if (left == -1) return right;
    if (right == -1) return left;
    return left < right ? left : right;
}

int findSecondMinimumValue(struct TreeNode* root) {
    if (!root) return -1;
    return dfs(root, root->val);
}