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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:
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
我以前为啥要做那么复杂呢
ref http://www.cnblogs.com/springfor/p/3889327.html
关键在于找到有效的下家P
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.right!=null) root.right.next = p; if(root.left!=null){ if(root.right!=null){ root.left.next = root.right; }else root.left.next =p; } connect(root.right); connect(root.left); }
Populating Next Right Pointers in Each Node I or II
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原文地址：http://www.cnblogs.com/jiajiaxingxing/p/4565358.html
