116. Populating Next Right Pointers in Each Node**

Given a binary tree

struct TreeLinkNode {
TreeLinkNode *left;
TreeLinkNode *right;
TreeLinkNode *next;
}

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to 

NULL

.

Initially, all next pointers are set to 

NULL

.

Note:
You may only use constant extra space.
You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

For example,

Given the following perfect binary tree,

1
/  \
2    3
/ \  / \
4  5  6  7

After calling your function, the tree should look like:

1 -> NULL
/  \
2 -> 3 -> NULL
/ \  / \
4->5->6->7 -> NULL

My code:

/**
* 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) {
Queue<TreeLinkNode> queue = new LinkedList<>();
if(root==null) return;
queue.offer(root);
while(!queue.isEmpty()){
int qSize = queue.size();
for (int i=0;i<qSize;i++){
TreeLinkNode cur = queue.poll();
if(i!=qSize-1) cur.next = queue.peek();
else cur.next = null;
if (cur.left!=null) queue.offer(cur.left);
if (cur.right!=null) queue.offer(cur.right);
}
}
}
}

总结：你是最棒的，不急不躁，慢慢来。加油啦。一切都会好的。

O(n)space，O(n)time

改进：

public class Solution {
public void connect(TreeLinkNode root) {
TreeLinkNode level_start=root;
while(level_start!=null){
TreeLinkNode cur=level_start;
while(cur!=null){
if(cur.left!=null) cur.left.next=cur.right;
if(cur.right!=null && cur.next!=null) cur.right.next=cur.next.left;

cur=cur.next;
}
level_start=level_start.left;
}
}
}

O(1)space, O(n)time