leetcode116~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

public class PopulatingNextRightPointers116 {
//通用的解法，适用于所有二叉树
public void connect(TreeLinkNode root) {
if(root==null) return;
Queue<TreeLinkNode> queue = new LinkedList<>();
queue.offer(root);
while(!queue.isEmpty()) {
int size = queue.size();
TreeLinkNode tmp = null;
for(int i=0;i<size;i++) {
TreeLinkNode node1 = queue.poll();
if(node1.left!=null) {
queue.offer(node1.left);
}
if(node1.right!=null) {
queue.offer(node1.right);
}
if(tmp!=null) {
tmp.next=node1;
}
if(i==size-1) {
node1.next=null;
tmp=null;
continue;
}
tmp = node1;
}
}
}

//通用的解法，适用于所有二叉树
//与上一方法类似
public void connect3(TreeLinkNode root) {
if(root==null) return;
Queue<TreeLinkNode> queue = new LinkedList<TreeLinkNode>();
queue.offer(root);
//记录每层节点的总数
int numOfLevelNode = 1;
while(!queue.isEmpty()) {
TreeLinkNode node = queue<
a423
/span>.poll();
numOfLevelNode--;
if(node.left!=null) {
queue.offer(node.left);
}
if(node.right!=null) {
queue.offer(node.right);
}
if(numOfLevelNode>0) {
node.next=queue.peek();
} else {
numOfLevelNode=queue.size();
}
}
}

/*
*  以下解法只适应于完全二叉树
如果当前节点有左孩子，那么左孩子的next就指向右孩子；如果当前节点有右孩子，
那么判断，如果当前节点的next是null，则说明当前节点已经在最右边，那么右孩子也是最右边，
所以右孩子指向null；如果当前节点的next不是null，那么当前节点的右孩子的next需要指向当前节点
next的左孩子
*/
public void connect2(TreeLinkNode root) {
if(root==null) return;
if(root.left!=null) {
root.left.next = root.right;
}
if(root.right!=null) {
if(root.next==null) {
root.right.next=null;
} else {
root.right.next=root.next.left;
}
}
connect2(root.left);
connect2(root.right);
}
}