Given a non-empty special binary tree consisting of nodes with the non-negative value, where each node in this tree has exactly twoor 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.
class Solution {
public:
int findSecondMinimumValue(TreeNode* root) {
if (!root) return -1;
int ans = minval(root, root->val);
return ans;
}
private:
int minval(TreeNode* p, int first) {
if (p == nullptr) return -1;
if (p->val != first) return p->val;
int left = minval(p->left, first), right = minval(p->right, first);
// if all nodes of a subtree = root->val,
// there is no second minimum value, return -1
if (left == -1) return right;
if (right == -1) return left;
return min(left, right);
}
};
二叉树第二小值
本文介绍了一种算法,用于找到特殊二叉树中节点值的第二最小值。这种二叉树每个节点的值等于其两个子节点中的较小值。文章提供了一个具体的实现示例,并通过两个例子展示了如何寻找第二最小值。

被折叠的 条评论
为什么被折叠?



