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
代码如下:
/**
* 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) {
int a=1;
int b=a;
List<TreeLinkNode> list=LevelTraverse(root); try{
int i=0;
while(i<list.size())
{
TreeLinkNode node=list.get(i);
i++;
b=b-1;
while(b>0)
{
node.next=list.get(i);
i++;
node=node.next;
b--;
}
node.next=null;
b=a*2;
a=b;
}
}catch(NullPointerException e){} }
public ArrayList<TreeLinkNode> LevelTraverse(TreeLinkNode root)//树的水平遍历
{
ArrayList<TreeLinkNode> list=new ArrayList<TreeLinkNode>();
Queue<TreeLinkNode> queue = new LinkedList<TreeLinkNode>();
queue.add(root);
try{
while(queue.size()>0)
{
TreeLinkNode a=(TreeLinkNode)queue.peek();
queue.remove();
list.add(a);
if(a.left!=null)
queue.add(a.left);
if(a.right!=null)
queue.add(a.right);
}
}catch(NullPointerException e){} return list;
}
}
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