[Leetcode] 671. Second Minimum Node In a Binary Tree 解题报告

最新推荐文章于 2021-08-07 12:17:51 发布

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本文探讨了如何在特殊二叉树中寻找第二最小值的问题，该二叉树的每个节点值等于其两个子节点中的较小值。文章通过递归算法解决此问题，并提供了具体的实现代码。

题目：

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.

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: 
    2
   / \
  2   5
     / \
    5   7

Output: 5
Explanation: The smallest value is 2, the second smallest value is 5.

Example 2:

Input: 
    2
   / \
  2   2

Output: -1
Explanation: The smallest value is 2, but there isn't any second smallest value.

思路：

我们首先计算出根节点给的值，作为参考。然后递归执行，计算左子树和右子树的值，如果两者都比根节点的值大，那么取其小者；否则取其大者（因为此时至少有一个是-1

）。

代码：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int findSecondMinimumValue(TreeNode* root) {
        if (!root) {
            return -1;
        }
        return minVal(root, root->val);
    }
private:
    int minVal(TreeNode* p, int value) {
        if (p == NULL) {
            return -1;
        }
        if (p->val != value) {                  // different from current minimum value
            return p->val;
        }
        int left = minVal(p->left, value);
        int right = minVal(p->right, value);
        if (left == -1 || right == -1) {        // at least one of them has no value larger than value
            return max(left, right);
        }
        else {                                  // both of them has larger value
            return min(left, right);
        }
    }
};