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) { //注意: //1、默认next为空,因此不用讨论最右侧节点的next问题(包括根节点) //2、注意例子中5->6,求非最外侧的右节点的next,此时需要结合根节点是否有next进行判断 if(root==null) return; if(root.left!=null){ root.left.next=root.right; } if(root.right!=null){ if(root.next!=null){ root.right.next=root.next.left; } } connect(root.left); connect(root.right); } }