111.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

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分析：给定一个完全二叉树，给每个结点加上其可以指向所在层的右边结点，没有的话则指向null。
首先按层找到每一层的结点，然后让每个结点指向其右边的结点。

/**
* 给定一个完全二叉树，给每个结点加上其可以指向所在层的右边结点，没有的话则指向null。
* Step1:按层获取每层的结点；
* Step2:每层的结点right指向其右边结点，最右边指向null。
* @date 20160417
*/
public void connect(TreeLinkNode root) {
/*如果根结点为空则直接返回*/
if(root == null){
return;
}
List<TreeLinkNode> currentlist = new ArrayList<TreeLinkNode>();//存放当前层的结点，初始化时放入根结点
List<TreeLinkNode> nextlist = new ArrayList<TreeLinkNode>();//当前层的孩子结点层
currentlist.add(root);
TreeLinkNode node ;
while(currentlist.size()!=0){
int size = currentlist.size();
/*依次遍历当前层中每个结点，并把拿到的结点的next指向其下一个结点，同时把当前结点的孩子结点放到孩子层*/
int i=0;
for(;i<size-1;i++){
node = currentlist.get(i);
node.next = currentlist.get(i+1);
if(node.left!=null){
nextlist.add(node.left);
}
if(node.right!=null){
nextlist.add(node.right);
}
}
node=currentlist.get(i);
node.next =null;//当前层的最右边指向null
if(node.left!=null){
nextlist.add(node.left);
}
if(node.right!=null){
nextlist.add(node.right);
}

currentlist.clear();
currentlist.addAll(nextlist);
nextlist.clear();

}//end while
}