题目:
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 题解:
这道题跟I的区别就是binary tree不是完全二叉树。
所以root.right.next就不一定等于root.next.left。
所以,目标就是先确定好root的右孩子的第一个有效next连接点,然后再处理左孩子。 代码如下:
1 public void connect(TreeLinkNode root) {
2 if (root == null)
3 return;
4
5 TreeLinkNode p = root.next;
6 /*
7 因此,这道题目首要是找到右孩子的第一个有效的next链接节点,然后再处理左孩子。然后依次递归处理右孩子,左孩子
8 */
9 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);
}