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.
- Recursive approach is fine, implicit stack space does not count as extra space for this problem.
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
思路:基于层次遍历,用已构建的链表用于保存下一层的结构信息。
/** * 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; while(root != null){ TreeLinkNode head = new TreeLinkNode(0); TreeLinkNode temp = head; while( root !=null ){ if(root.left != null){ temp.next = root.left; temp = temp.next; } if(root.right != null){ temp.next = root.right; temp = temp.next; } root = root.next; } root = head.next; } } }