Follow up for problem "Populating Next Right Pointers in Each Node". What if the given tree could be any binary tree? Would your previous solution still work? Note: You may only use constant extra space. For example, Given the following binary tree, 1 / 2 3 / 4 5 7 After calling your function, the tree should look like: 1 -> NULL / 2 -> 3 -> NULL / 4-> 5 -> 7 -> NULL
1 class Solution { 2 public Node connect(Node root) { 3 Node pre = null; 4 if (root == null) return null; 5 Queue<Node> queue = new LinkedList<>(); 6 queue.offer(root); 7 while (!queue.isEmpty()) { 8 int size = queue.size(); 9 for (int i = 0; i < size; i++) { 10 Node cur = queue.poll(); 11 if (pre == null) pre = cur; 12 else { 13 pre.next = cur; 14 pre = cur; 15 } 16 if (cur.left != null) queue.offer(cur.left); 17 if (cur.right != null) queue.offer(cur.right); 18 } 19 pre = null; 20 } 21 return root; 22 } 23 }
层次递进法
复杂度
时间 O(N) 空间 O(1)
1 public class Solution { 2 3 //based on level order traversal 4 public void connect(TreeLinkNode root) { 5 6 TreeLinkNode head = null; //head of the next level 7 TreeLinkNode prev = null; //the leading node on the next level 8 TreeLinkNode cur = root; //current node of current level 9 10 while (cur != null) { 11 12 while (cur != null) { //iterate on the current level 13 //left child 14 if (cur.left != null) { 15 if (prev != null) { 16 prev.next = cur.left; 17 } else { 18 head = cur.left; 19 } 20 prev = cur.left; 21 } 22 //right child 23 if (cur.right != null) { 24 if (prev != null) { 25 prev.next = cur.right; 26 } else { 27 head = cur.right; 28 } 29 prev = cur.right; 30 } 31 //move to next node 32 cur = cur.next; 33 } 34 35 //move to next level 36 cur = head; 37 head = null; 38 prev = null; 39 } 40 41 } 42 }