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 }
Leetcode: Populating Next Right Pointers in Each Node II
原文:http://www.cnblogs.com/EdwardLiu/p/3978460.html