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
以下转自:
http://www.cnblogs.com/feiling/p/3268836.html
对于任意的二叉树,上题的解不能解决问题,主要原因在于上题有效的next都是直接的sibling
而本题有效的next肯能在不同的子树上,另外本题需要先递归处理右子树,再处理左子树
如下面的例子,如果先处理左子树,当处理到节点7的右子树节点0,我们发现节点7的next是9,但9没有子树,
这时候我们应该去看9的next,但此时右子树并没有进行处理,9的next为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; TreeLinkNode p = root.next; while(p != null){ if(p.left != null){ p = p.left; break; } if(p.right != null){ p = p.right; break; } p = p.next; } if(root.left != null){ if(root.right == null) root.left.next = p; else root.left.next = root.right; } if(root.right != null){ if(root.next == null) root.right.next = null; else root.right.next = p; } connect(root.right); connect(root.left); } }