Leetcode: Populating Next Right Pointers in Each Node II

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

在Populating Next Right Pointers in Each Node问题的基础上，难度20，方法一样。都是类似Binary Tree Level Order Traverse，都是把树看成一个无向图，然后用BFS的方式，需要记录每一层的ParentNumInQueue以及ChildNumInQueue, 初始值为1和0，以后每次ParentNumInQ减至0说明这一层已经遍历完毕，这一层的Child数将成为下一层的ParentNumInQ

 1 /**
 2  * Definition for binary tree with next pointer.
 3  * public class TreeLinkNode {
 4  *     int val;
 5  *     TreeLinkNode left, right, next;
 6  *     TreeLinkNode(int x) { val = x; }
 7  * }
 8  */
 9 public class Solution {
10     public void connect(TreeLinkNode root) {
11         if (root == null) return;
12         LinkedList<TreeLinkNode> queue = new LinkedList<TreeLinkNode>();
13         queue.add(root);
14         int ParentNumInQ = 1;
15         int ChildNumInQ = 0;
16         TreeLinkNode pre = null;
17         while (!queue.isEmpty()) {
18             TreeLinkNode cur = queue.poll();
19             ParentNumInQ--;
20             if (pre == null) {
21                 pre = cur;
22             }
23             else {
24                 pre.next = cur;
25                 pre = pre.next;
26             }
27             if (cur.left != null) {
28                 queue.add(cur.left);
29                 ChildNumInQ++;
30             }
31             if (cur.right != null) {
32                 queue.add(cur.right);
33                 ChildNumInQ++;
34             }
35             if (ParentNumInQ == 0) {
36                 ParentNumInQ = ChildNumInQ;
37                 ChildNumInQ = 0;
38                 pre.next = null;
39                 pre = null;
40             }
41         }
42     }
43 }

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 }