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
一刷
题解：
创建3个节点，cur(处于当前层), prev(处于操作层), head(下一层的最外节点)。即 cur-1, prev-2, head-3
/**
* 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) {
TreeLinkNode head = null;//the head of the next level
TreeLinkNode prev = null;// the leading node of next level
TreeLinkNode cur = root; //current node of current level
//cur-1, prev-2, head-3
while(cur!=null){
while(cur!=null){
//current manipulate-level 2
if(cur.left!=null){
if(prev == null){
head = cur.left;
}
else prev.next = cur.left;
prev = cur.left;
}
if(cur.right!=null){
if(prev == null) head = cur.right;
else prev.next = cur.right;
prev = cur.right;
}
cur = cur.next;
}
cur = head;
head = null;
prev = null;
}
}
}
