[2018-Interview] Binary Tree Level Order Traversal

Original question: https://leetcode.com/problems/binary-tree-level-order-traversal/description/

My Answer:

Iteration:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> allNodes = new ArrayList<List<Integer>>();
        if (root == null) {
            return allNodes;
        }
        List<TreeNode> currentLevel = new ArrayList<TreeNode>();
        currentLevel.add(root);
        int currentLevelSize = currentLevel.size();
        int nextLevelSize = 0;
        while(!currentLevel.isEmpty()) {
            List<Integer> values = new ArrayList<Integer>();
            while (currentLevelSize > 0) {
                TreeNode node = currentLevel.get(0);
                currentLevel.remove(0);
                values.add(node.val);
                if (node.left != null) {
                    currentLevel.add(node.left);
                    nextLevelSize ++;
                }
                if (node.right != null) {
                    currentLevel.add(node.right);
                    nextLevelSize ++;
                }
                currentLevelSize --;
            }
            allNodes.add(values);
            currentLevelSize = nextLevelSize;
            nextLevelSize = 0;
        }
        
        return allNodes;
    }
}

Recursion:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> allNodes = new ArrayList<List<Integer>>();
        if (root == null) {
            return allNodes;
        }
        helper(root, 0, allNodes);
        
        return allNodes;
    }

    void helper(TreeNode node, int level, List<List<Integer>> allNodes) {
        if (node == null) {
            return;
        }

        List<Integer> values = new ArrayList<Integer>();
        values.add(node.val);

        if (allNodes.size() > level) {
            List<Integer> existingValues = allNodes.get(level);
            existingValues.addAll(values);
            allNodes.remove(level);
            allNodes.add(level, existingValues);
        } else {
            allNodes.add(values);
        }
        
        helper(node.left, level + 1, allNodes);
        helper(node.right, level + 1, allNodes);
    }
}

[2018-Interview] Symmetric Tree

Original question: https://leetcode.com/problems/symmetric-tree/description/

My answer:

Recursive:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root == null) {
            return true;
        }
        return helper(root.left, root.right);
    }
    
    boolean helper(TreeNode left, TreeNode right) {
        if (left == null && right == null) {
            return true;
        }

        if ((left == null && right != null) ||
            (left != null && right == null)
           ) {
            return false;
        } else {
            // left != null && right != null
            if (left.val == right.val) {
                return helper(left.left, right.right) && helper(left.right, right.left);   
            } else {
                return false;
            }
        }
    }
}

Iterative:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root == null) {
            return true;
        }
        LinkedList<TreeNode> currentLevel = new LinkedList<TreeNode>();
        LinkedList<TreeNode> nextLevel = new LinkedList<TreeNode>();
        currentLevel.addLast(root);
        while (true) {
            while(!currentLevel.isEmpty()) {
                TreeNode temp = currentLevel.removeFirst();
                if (temp != null) {
                    nextLevel.addLast(temp.left);
                    nextLevel.addLast(temp.right);
                }
            }
            if (nextLevel.isEmpty()) {
                break;
            }
            int left = (nextLevel.size() - 1) / 2 ;
            int right = left + 1;
            while (left >= 0 && right < nextLevel.size()) {
                if (!equals(nextLevel.get(left), nextLevel.get(right))) {
                    return false;
                }
                left --;
                right ++;
            }
            currentLevel.addAll(nextLevel);
            nextLevel = new LinkedList<TreeNode>();
        }
        return true;
    }

    boolean equals(TreeNode node1, TreeNode node2) {
        if (node1 == null && node2 == null) {
            return true;
        } else {
            if (node1 != null && node2 != null && node1.val == node2.val) {
                return true;
            }
        }
        return false;
    }
}

[2018-Interview] Maximum Depth of Binary Tree

Original question: https://leetcode.com/problems/maximum-depth-of-binary-tree/description/

My answer: we can have 2 solutions: recursive and iterative. Solution 2 uses 2 queues to keep track of current level nodes and next level nodes.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int maxDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        // Solution 1: recursive
//         int leftMaxDepth = maxDepth(root.left);
//         int rightMaxDepth = maxDepth(root.right);
//         return (Math.max(leftMaxDepth, rightMaxDepth) + 1);
        
        // Solution 2: iterative
        int maxDepth = 1;
        LinkedList<TreeNode> currentLevel = new LinkedList<TreeNode>();
        LinkedList<TreeNode> nextLevel = new LinkedList<TreeNode>();
        currentLevel.add(root);
        while(true) {
            while (!currentLevel.isEmpty()) {
                TreeNode currentNode = currentLevel.removeFirst();
                if (currentNode.left != null) {
                    nextLevel.add(currentNode.left);
                }
                if (currentNode.right != null) {
                    nextLevel.add(currentNode.right);
                }
            }
            if (nextLevel.isEmpty()) {
                break;
            }
            maxDepth ++;
            currentLevel.addAll(nextLevel);
            nextLevel = new LinkedList<TreeNode>();
        }
        return maxDepth;
    }
}

[2018-Interview] Linked List Cycle

Original question: https://leetcode.com/problems/linked-list-cycle/description/

My answer: have 2 nodes running in the list, one fast, one slow. If there is a cycle, they will meet somewhere. If not, the fast will reach end of the list first.

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public boolean hasCycle(ListNode head) {
        if (head == null) {
            return false;
        }
        ListNode p1 = head;
        ListNode p2 = head.next;

        while(p2 != null && p2.next != null && p1 != p2) {
            p1 = p1.next;
            p2 = p2.next.next;
        }

        if (p2 == p1) {
            return true;
        }
        return false;
    }
}

[2018-Interview] Palindrome Linked List

Original question: https://leetcode.com/problems/palindrome-linked-list/description/

Answer: Basic logic, find the middle node, and reverse the right-side nodes, then compare left-side nodes with right-side nodes. Answer is from Jiuzhang: https://www.jiuzhang.com/solution/palindrome-linked-list/

/**
* 本参考程序来自九章算法，由 @九章算法 提供。版权所有，转发请注明出处。
* - 九章算法致力于帮助更多中国人找到好的工作，教师团队均来自硅谷和国内的一线大公司在职工程师。
* - 现有的面试培训课程包括：九章算法班，系统设计班，算法强化班，Java入门与基础算法班，Android 项目实战班，
* - Big Data 项目实战班，算法面试高频题班, 动态规划专题班
* - 更多详情请见官方网站：http://www.jiuzhang.com/?source=code
*/ 

// This code would destroy the original structure of the linked list.
// If you do not want to destroy the structure, you can reserve the second part back.
public class Solution {
    /**
     * @param head a ListNode
     * @return a boolean
     */
    public boolean isPalindrome(ListNode head) {
        if (head == null) {
            return true;
        }
        
        ListNode middle = findMiddle(head);
        middle.next = reverse(middle.next);
        
        ListNode p1 = head, p2 = middle.next;
        while (p1 != null && p2 != null && p1.val == p2.val) {
            p1 = p1.next;
            p2 = p2.next;
        }
        
        return p2 == null;
    }
    
    private ListNode findMiddle(ListNode head) {
        if (head == null) {
            return null;
        }
        ListNode slow = head, fast = head.next;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        
        return slow;
    }
    
    private ListNode reverse(ListNode head) {
        ListNode prev = null;
        
        while (head != null) {
            ListNode temp = head.next;
            head.next = prev;
            prev = head;
            head = temp;
        }
        
        return prev;
    }
}

[2018-Interview] Merge Two Sorted Lists

Original question: https://leetcode.com/problems/merge-two-sorted-lists/description/

My answer: pretty simple and straightforward.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null) {
            return l2;
        } else if (l2 == null) {
            return l1;
        }

        ListNode head = new ListNode(-1);        
        ListNode current = head;

        while(l1 != null && l2 != null) {
            if (l1.val > l2.val) {
                current.next = l2;
                l2 = l2.next;
            } else {
                current.next = l1;
                l1 = l1.next;
            }
            current = current.next;
        }
        ListNode remainingNode = l1 == null? l2 : l1;
        current.next = remainingNode;
        return head.next;
    }
}