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:
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Input: 2 / \ 2 5 / \ 5 7 Output: 5 Explanation: The smallest value is 2, the second smallest value is 5. |
Example 2:
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Input: 2 / \ 2 2 Output: -1 Explanation: The smallest value is 2, but there isn't any second smallest value. |
Solution:
Time complexity O(n), Space complexity O(n)
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/** * 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 BFS(root); return DFS(root, root->val); } private: // Time complexity: O(n) // Space complexity: O(n) int BFS(TreeNode* root) { if(!root) return -1; // Smallest value int s1 = root->val; // Sencond smallest value int s2 = INT_MAX; bool found = false; deque<TreeNode*> q; q.push_back(root); while(!q.empty()) { TreeNode* node = q.front(); q.pop_front(); // Keep updating second smallest value if(node->val > s1 && node->val < s2) { s2 = node->val; found = true; // No need to add it's children into queue continue; } if(!node->left) continue; q.push_back(node->left); q.push_back(node->right); } return found?s2:-1; } // Return the second smallest number in the (sub-tree) // s1 is the smallest value int DFS(TreeNode* root, int s1) { if(!root) return -1; // If root's value is already grater than s1, // then all its children's value should be >= s1. // Thus root's value is the second smallest one. // No need to visit the if(root->val > s1) return root->val; int sl = DFS(root->left, s1); int sr = DFS(root->right, s1); if(sl == -1) return sr; if(sr == -1) return sl; // Return the smaller one among two subtrees return min(sl, sr); } }; |
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