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【LeetCode OJ】Populating Next Right Pointers in Nod

2014-06-09 00:00 232 查看
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


/**
* 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) {
if(root == null)return;
Queue<TreeLinkNode> q = new LinkedList<TreeLinkNode>();
q.add(root);
int n = 0;
TreeLinkNode temp,head = new TreeLinkNode(0);
while(!q.isEmpty()){
temp = q.poll();
if(temp.left != null){
if(q.size() == Math.pow(2, n) - 1){
++n;
root.next = null;
head = temp.left;
head.next = temp.right;
head = temp.right;
}else{
head.next = temp.left;
temp.left.next = temp.right;
head = temp.right;
}
q.add(temp.left);
q.add(temp.right);
}
}
}
}
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