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Formatted question description: https://leetcode.ca/all/671.html
671. Second Minimum Node In a Binary Tree (Easy)
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 <= 231 - 1root.val == min(root.left.val, root.right.val)for each internal node of the tree.
Related Topics:
Tree
Similar Questions:
Solution 1. DFS
// OJ: https://leetcode.com/problems/second-minimum-node-in-a-binary-tree/
// Time: O(N)
// Space: O(H)
class Solution {
long mn, ans = LONG_MAX;
void dfs(TreeNode *root) {
if (root->left) {
if (root->left->val == mn) dfs(root->left);
else ans = min(ans, (long)root->left->val);
}
if (root->right) {
if (root->right->val == mn) dfs(root->right);
else ans = min(ans, (long)root->right->val);
}
}
public:
int findSecondMinimumValue(TreeNode* root) {
mn = root->val;
dfs(root);
return ans == LONG_MAX ? -1 : ans;
}
};
Or
// OJ: https://leetcode.com/problems/second-minimum-node-in-a-binary-tree/
// Time: O(N)
// Space: O(H)
class Solution {
long mn, ans = LONG_MAX;
void dfs(TreeNode *root) {
if (!root) return;
if (root->val > mn && root->val < ans) ans = min(ans, (long)root->val);
if (root->val == mn) {
dfs(root->left);
dfs(root->right);
}
}
public:
int findSecondMinimumValue(TreeNode* root) {
mn = root->val;
dfs(root);
return ans == LONG_MAX ? -1 : ans;
}
};
-
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public int findSecondMinimumValue(TreeNode root) { List<Integer> values = new ArrayList<Integer>(); Queue<TreeNode> queue = new LinkedList<TreeNode>(); queue.offer(root); while (!queue.isEmpty()) { TreeNode node = queue.poll(); int value = node.val; if (!values.contains(value)) values.add(value); TreeNode left = node.left, right = node.right; if (left != null) queue.offer(left); if (right != null) queue.offer(right); } Collections.sort(values); if (values.size() < 2) return -1; else return values.get(1); } } ############ /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * this.left = left; * this.right = right; * } * } */ class Solution { private int ans = -1; public int findSecondMinimumValue(TreeNode root) { dfs(root, root.val); return ans; } private void dfs(TreeNode root, int val) { if (root != null) { dfs(root.left, val); dfs(root.right, val); if (root.val > val) { ans = ans == -1 ? root.val : Math.min(ans, root.val); } } } } -
// OJ: https://leetcode.com/problems/second-minimum-node-in-a-binary-tree/ // Time: O(N) // Space: O(H) class Solution { long mn, ans = LONG_MAX; void dfs(TreeNode *root) { if (root->left) { if (root->left->val == mn) dfs(root->left); else ans = min(ans, (long)root->left->val); } if (root->right) { if (root->right->val == mn) dfs(root->right); else ans = min(ans, (long)root->right->val); } } public: int findSecondMinimumValue(TreeNode* root) { mn = root->val; dfs(root); return ans == LONG_MAX ? -1 : ans; } }; -
# Definition for a binary tree node. # class TreeNode: # def __init__(self, val=0, left=None, right=None): # self.val = val # self.left = left # self.right = right class Solution: def findSecondMinimumValue(self, root: Optional[TreeNode]) -> int: def dfs(root): if root: dfs(root.left) dfs(root.right) nonlocal ans, v if root.val > v: ans = root.val if ans == -1 else min(ans, root.val) ans, v = -1, root.val dfs(root) return ans ############ # Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution(object): def findSecondMinimumValue(self, root): """ :type root: TreeNode :rtype: int """ self.res = set() self.inOrder(root) if len(self.res) <= 1: return -1 min1 = min(self.res) self.res.remove(min1) return min(self.res) def inOrder(self, root): if not root: return self.inOrder(root.left) self.res.add(root.val) self.inOrder(root.right) -
/** * Definition for a binary tree node. * type TreeNode struct { * Val int * Left *TreeNode * Right *TreeNode * } */ func findSecondMinimumValue(root *TreeNode) int { ans, v := -1, root.Val var dfs func(*TreeNode) dfs = func(root *TreeNode) { if root == nil { return } dfs(root.Left) dfs(root.Right) if root.Val > v { if ans == -1 || ans > root.Val { ans = root.Val } } } dfs(root) return ans } -
/** * Definition for a binary tree node. * function TreeNode(val, left, right) { * this.val = (val===undefined ? 0 : val) * this.left = (left===undefined ? null : left) * this.right = (right===undefined ? null : right) * } */ /** * @param {TreeNode} root * @return {number} */ var findSecondMinimumValue = function (root) { let ans = -1; const v = root.val; function dfs(root) { if (!root) { return; } dfs(root.left); dfs(root.right); if (root.val > v) { if (ans == -1 || ans > root.val) { ans = root.val; } } } dfs(root); return ans; };