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.
Definition For a Binary Free
function TreeNode(val) {
this.val = val;
this.left = this.right = null;
}
Solution (DFS)
/**
* Second Minimum Node In a Binary Tree
* @param {TreeNode} root
* @return {number}
* @time complexity: O(n) where n is the total number of nodes
* @space complexity: O(n) the data stored in min and res are O(1),
* but the depth first search may take up O(h) in stack call,
* where h is the height of the tree
*/
var findSecondMinimumValue = function(root) {
let min = root.val;
let res = Infinity;
const dfs = function(node) {
if (!node) {
return;
}
if (node.val === min) {
dfs(node.left);
dfs(node.right);
} else if (node.val > min && node.val < res) {
res = node.val;
}
}
dfs(root);
return res === Infinity ? -1 : res;
};
Solution (BFS)
/**
* Second Minimum Node In a Binary Tree
* @param {TreeNode} root
* @return {number}
* @time complexity: O(n) where n is the total number of nodes
* @space complexity: O(n)
*/
var findSecondMinimumValue = function(root) {
const stack = [root];
let min = root.val;
let res = Infinity;
while (stack.length) {
const node = stack.shift();
if (node.val === min && node.left) {
stack.push(node.left);
stack.push(node.right);
} else if (node.val > min && node.val < res) {
res = node.val;
}
}
return res === Infinity ? -1 : res;
};
Alternative Solution
/**
* Second Minimum Node In a Binary Tree
* @param {TreeNode} root
* @return {number}
* @time complexity: O(n) where n is the total number of nodes
* @space complexity: O(n) the data stored in min and res are O(1),
* but the depth first search may take up O(h) in stack call,
* where h is the height of the tree
*/
var findSecondMinimumValue = function(root) {
let first = Infinity;
let second = Infinity;
const bfs = function(root) {
if (!root) {
return;
}
if (root.val < first && root.val !== second) {
second = first;
first = root.val;
} else if (root.val < second && root.val !== first) {
second = root.val;
}
bfs(root.left);
bfs(root.right);
}
bfs(root);
return second === Infinity ? -1 : second;
}
Test Case
const assert = require('chai').assert;
const arrayToTree = function(arr, root = null, index = 0) {
if (index < arr.length) {
root = new TreeNode(arr[index]);
root.left = arrayToTree(arr, root.left, 2 * index + 1);
root.right = arrayToTree(arr, root.right, 2 * index + 2);
}
return root;
}
describe('Second Minimum Node In a Binary Tree', () => {
it('should return the second minimum node', () => {
const tree = arrayToTree([2,2,5,null,null,5,7]);
assert.strictEqual(findSecondMinimumValue(tree), 5);
});
it('should return -1 when there is no such second minimum node', () => {
const tree = arrayToTree([2,2,2]);
assert.strictEqual(findSecondMinimumValue(tree), -1);
});
});
Second Minimum Node In a Binary Tree
✓ should return the second minimum node
✓ should return -1 when there is such second minimum node
2 passing (12ms)
