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/*******************************************************************************
* Setup
******************************************************************************/
interface BinaryTreeNode {
value: number;
left: BinaryTreeNode;
right: BinaryTreeNode;
}
class BinaryTreeNode {
constructor(value: number) {
this.value = value;
this.left = null;
this.right = null;
}
insertLeft(value: number): BinaryTreeNode {
this.left = new BinaryTreeNode(value);
return this.left;
}
insertRight(value: number): BinaryTreeNode {
this.right = new BinaryTreeNode(value);
return this.right;
}
}
/*******************************************************************************
* First solution
******************************************************************************/
/**
* I first solved this problem using O(n) space and O(n*log(n))
* time. InterviewCake informs me that we can improve both the time and the
* space performance.
*/
function findSecondLargest_first(node: BinaryTreeNode): number {
const stack: Array<BinaryTreeNode> = [];
const xs: Array<number> = [];
stack.push(node);
while (stack.length > 0) {
const node = stack.pop()
xs.push(node.value);
if (node.left) {
stack.push(node.left);
}
if (node.right) {
stack.push(node.right);
}
}
xs.sort();
if (xs.length < 2) {
throw new Error('Cannot find the second largest element in a BST with fewer than two elements.');
} else {
return xs[xs.length - 2];
}
}
/*******************************************************************************
* Second solution
******************************************************************************/
/**
* My second solution accumulates a list of the values in the tree using an
* in-order traversal. This reduces the runtime costs from O(n*log(n)) from the
* previous solution to O(n). The memory cost is still O(n), which InterviewCake
* informs me can be reduced to O(1).
*/
function findSecondLargest_second(node: BinaryTreeNode): number {
const xs: Array<number> = accumulateInorder(node);
if (xs.length < 2) {
throw new Error('Cannot find the second largest element in a BST with fewer than two elements.');
} else {
return xs[xs.length - 2];
}
}
/**
* Returns an array containing the values of the tree, `node`, sorted in-order
* (i.e. from smallest-to-largest).
*/
function accumulateInorder(node: BinaryTreeNode): Array<number> {
let result = [];
if (node.left) {
result = result.concat(accumulateInorder(node.left));
}
result.push(node.value)
if (node.right) {
result = result.concat(accumulateInorder(node.right));
}
return result;
}
/*******************************************************************************
* Third solution
******************************************************************************/
/**
* Returns the largest number in a BST.
*/
function findLargest(node: BinaryTreeNode): number {
let curr: BinaryTreeNode = node;
while (curr.right) {
curr = curr.right;
}
return curr.value;
}
/**
* Returns the second largest number in a BST
*/
function findSecondLargest(node: BinaryTreeNode): number {
let curr = node;
let parent = null;
while (curr.right) {
parent = curr;
curr = curr.right
}
if (curr.left) {
return findLargest(curr.left);
}
else {
return parent.value;
}
}
// Tests
let desc = 'full tree';
let treeRoot = new BinaryTreeNode(50);
let leftNode = treeRoot.insertLeft(30);
leftNode.insertLeft(10);
leftNode.insertRight(40);
let rightNode = treeRoot.insertRight(70);
rightNode.insertLeft(60);
rightNode.insertRight(80);
assertEquals(findSecondLargest(treeRoot), 70, desc);
desc = 'largest has a left child';
treeRoot = new BinaryTreeNode(50);
leftNode = treeRoot.insertLeft(30);
leftNode.insertLeft(10);
leftNode.insertRight(40);
rightNode = treeRoot.insertRight(70);
rightNode.insertLeft(60);
assertEquals(findSecondLargest(treeRoot), 60, desc);
desc = 'largest has a left subtree';
treeRoot = new BinaryTreeNode(50);
leftNode = treeRoot.insertLeft(30);
leftNode.insertLeft(10);
leftNode.insertRight(40);
rightNode = treeRoot.insertRight(70);
leftNode = rightNode.insertLeft(60);
leftNode.insertRight(65);
leftNode = leftNode.insertLeft(55);
leftNode.insertRight(58);
assertEquals(findSecondLargest(treeRoot), 65, desc);
desc = 'second largest is root node';
treeRoot = new BinaryTreeNode(50);
leftNode = treeRoot.insertLeft(30);
leftNode.insertLeft(10);
leftNode.insertRight(40);
rightNode = treeRoot.insertRight(70);
assertEquals(findSecondLargest(treeRoot), 50, desc);
desc = 'descending linked list';
treeRoot = new BinaryTreeNode(50);
leftNode = treeRoot.insertLeft(40);
leftNode = leftNode.insertLeft(30);
leftNode = leftNode.insertLeft(20);
leftNode = leftNode.insertLeft(10);
assertEquals(findSecondLargest(treeRoot), 40, desc);
desc = 'ascending linked list';
treeRoot = new BinaryTreeNode(50);
rightNode = treeRoot.insertRight(60);
rightNode = rightNode.insertRight(70);
rightNode = rightNode.insertRight(80);
assertEquals(findSecondLargest(treeRoot), 70, desc);
desc = 'one node tree';
treeRoot = new BinaryTreeNode(50);
assertThrowsError(() => findSecondLargest(treeRoot), desc);
desc = 'when tree is empty';
treeRoot = null;
assertThrowsError(() => findSecondLargest(treeRoot), desc);
function assertEquals(a, b, desc) {
if (a === b) {
console.log(`${desc} ... PASS`);
} else {
console.log(`${desc} ... FAIL: ${a} != ${b}`)
}
}
function assertThrowsError(func, desc) {
try {
func();
console.log(`${desc} ... FAIL`);
} catch (e) {
console.log(`${desc} ... PASS`);
}
}
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