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
/** * 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; } Queue<TreeLinkNode> queue = new LinkedList<TreeLinkNode>(); queue.offer(root); TreeLinkNode pre = root; while (!queue.isEmpty()){ TreeLinkNode p = queue.poll(); pre.next = p; pre = p; if (p.left!=null){ queue.offer(p.left); } if (p.right!=null){ queue.offer(p.right); } } TreeLinkNode p =root; while (p!=null){ p.next = null; p = p.right; } } }