Leetcode: Populating Next Right Pointers in Each Node IIFollow 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 7After 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 }