Populating Next Right Pointers in Each Node *
Given a binary tree
struct TreeLinkNode { TreeLinkNode *left; TreeLinkNode *right; TreeLinkNode *next; }
Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL
.
Initially, all next pointers are set to NULL
.
Note:
- You may only use constant extra space.
- You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).
For example,
Given the following perfect binary tree,
1 / \ 2 3 / \ / \ 4 5 6 7
After calling your function, the tree should look like:
1 -> NULL / \ 2 -> 3 -> NULL / \ / \ 4->5->6->7 -> NULL
还是层序遍历的思想:
/** * Definition for binary tree with next pointer. * public class TreeLinkNode { * int val; * TreeLinkNode left, right, next; * TreeLinkNode(int x) { val = x; } * } */ public class Solution { public void connect(TreeLinkNode root) { if(root==null) return; List<TreeLinkNode> level_list = new ArrayList<TreeLinkNode>();//存储当前层节点 List<TreeLinkNode> tmp_list = new ArrayList<TreeLinkNode>();//存储下层节点 level_list.add(root); int flag = 1; while(flag==1) { flag = 0; tmp_list.clear(); for(int i=0;i<level_list.size();i++) { if(level_list.get(i).left!=null) { tmp_list.add(level_list.get(i).left); flag = 1; } if(level_list.get(i).right!=null) { tmp_list.add(level_list.get(i).right); flag = 1; } if(i==level_list.size()-1) level_list.get(i).next = null; else level_list.get(i).next = level_list.get(i+1); } level_list.clear(); level_list.addAll(tmp_list); } } }