The 2021-07-14 algorithm problem

1. Circular linked list

Set the speed pointer. If there is a ring, the slow pointer must catch up with the fast pointer; If there is no loop, the fast pointer becomes NULL

class Solution {
    public boolean hasCycle(ListNode head) {
        if(head==null||head.next==null) {return false;
        }
        ListNode slow = head;
        ListNode fast = head.next;
        while(fast.next ! =null&& fast ! = slow) {if (fast == null) {
                return false;
            }
            fast = fast.next.next;
            slow = slow.next;
        }
        if (fast == slow) {
            return true;
        }
        return false; }}Copy the code

2. Binary tree forward traversal

Ideas:…

class Solution {

    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> l = new ArrayList<>();
        preorder(root, l);
        return l;
    }

    public void preorder(TreeNode root, List<Integer> l) {
        if (root == null) {
            return; } l.add(root.val); preorder(root.left, l); preorder(root.right, l); }}Copy the code

3. Backorder traversal of binary tree

Ideas:…

class Solution {

    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> l = new ArrayList<>();
        postorder(root, l);
        return l;
    }
    public void postorder(TreeNode root, List<Integer> l) {
        if (root == null) {
            return; } postorder(root.left, l); postorder(root.right, l); l.add(root.val); }}Copy the code