Linked list
Code directly.
public static class ListNode{ int val; ListNode next; ListNode(int x) { val = x; }//Definition of linked list in Leetcode } private static ListNode createLinkedList(int[] arr) {//Input the input array into the linked list if (arr.length == 0) { return null; } ListNode head = new ListNode(arr[0]); ListNode current = head; for (int i = 1; i < arr.length; i++) {//process current.next = new ListNode(arr[i]); current = current.next; } return head; } private static void printLinkedList(ListNode head){//Print linked list results ListNode current = head; while (current!=null){ System.out.printf("%d -> ",current.val); current = current.next; } System.out.println("NULL"); } public static void main(String[] args) { int[] x = {1,2,3,4,5,6}; ListNode list = createLinkedList(x); printLinkedList(list1); }
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The printed result is:
tree
Code directly.
public static class TreeNode { int val; TreeNode left; TreeNode right; TreeNode(int x) { val = x; } } public static TreeNode constructTree(Integer[] nums){ if (nums.length == 0) return new TreeNode(0); Deque<TreeNode> nodeQueue = new LinkedList<>(); TreeNode root = new TreeNode(nums[0]); nodeQueue.offer(root); TreeNode cur; // Record the number of nodes in the current row (note that it is not necessarily the power of 2, but the number of non empty nodes in the previous row multiplied by 2) int lineNodeNum = 2; // Record the starting position of the number in the current row in the array int startIndex = 1; // Record the number of remaining elements in the array int restLength = nums.length - 1; while(restLength > 0) { // Only the last line can be dissatisfied, and the rest must be full // //If the number of input arrays is wrong, jump out of the program directly // if (restLength < lineNodeNum) { // System.out.println("Wrong Input!"); // return new TreeNode(0); // } for (int i = startIndex; i < startIndex + lineNodeNum; i = i + 2) { // It indicates that the number in nums has been used up. At this time, you should stop traversing and return to root directly if (i == nums.length) return root; cur = nodeQueue.poll(); if (nums[i] != null) { cur.left = new TreeNode(nums[i]); nodeQueue.offer(cur.left); } // As above, it means that the number in nums has been used up. At this time, the traversal should be stopped and root can be returned directly if (i + 1 == nums.length) return root; if (nums[i + 1] != null) { cur.right = new TreeNode(nums[i + 1]); nodeQueue.offer(cur.right); } } startIndex += lineNodeNum; restLength -= lineNodeNum; lineNodeNum = nodeQueue.size() * 2; } return root; } public static void preOrder(TreeNode root) {//Preorder arrangement if (root == null) return; System.out.print(root.val + " "); preOrder(root.left); preOrder(root.right); } public static void midOrder(TreeNode root) {//Ordinal alignment if (root == null) return; midOrder(root.left); System.out.print(root.val + " "); midOrder(root.right); } public static void aftOrder(TreeNode root) {//Post order arrangement if (root == null) return; aftOrder(root.left); aftOrder(root.right); System.out.print(root.val + " "); } public static void main(String[] args) { Integer[] nums = {1,2,2,3,3,3,3}; TreeNode tree=constructTree(nums); preOrder(tree); System.out.println(); midOrder(tree); System.out.println(); aftOrder(tree); }
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The printed result is:
I used the code of a blogger for reference, but I forgot which one. If the blogger saw it, comment on it, I'll mark it, MoMA da.
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