Populating Next Right Pointers in Each Node I or II
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
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;}elseroot.left.next =p;}connect(root.right);connect(root.left);}转载于:https://www.cnblogs.com/jiajiaxingxing/p/4565358.html
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