671. Second Minimum Node In a Binary Tree

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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 propertyroot.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:

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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:

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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.

Hints/Notes

  • binary tree

Solution

Language: C++

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/**
 * 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 findSecondMinimumValue(TreeNode* root) {
        if (!root->left && !root->right) {
            return -1;
        }

        int left = root->left->val, right = root->right->val;
        // the child's val is the minimum of the child tree
        // 1. if the child->val != root->val, then no need to dig further
        // 2. if the child->val == root->val, it indicates that there can be
        //    value larger than child->val
        // if the return value is -1:
        //     1. either the node doesn't have children
        //     2. all its children have the same value
        if (left == root->val) {
            left = findSecondMinimumValue(root->left);
        }

        if (right == root->val) {
            right = findSecondMinimumValue(root->right);
        }

        if (left == -1) {
            return right;
        }

        if (right == -1) {
            return left;
        }

        return min(left, right);
    }
};