Populating Next Right Pointers in Each Node II

标签：leetcode

Follow up for problem "Populating Next Right Pointers in Each Node".

What if the given tree could be any binary tree? Would your previous solution still work?

Note:

• You may only use constant extra space.

For example,
Given the following binary tree,

         1
       /        2    3
     / \        4   5    7

After calling your function, the tree should look like:

         1 -> NULL
       /        2 -> 3 -> NULL
     / \        4-> 5 -> 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) {
        Queue<TreeLinkNode> queue = new LinkedList<TreeLinkNode>();
        int level = 1;//记录需要出队的元素数，
        if(root == null ){
            return;
        }
        queue.offer(root);
        int count = 0;
        TreeLinkNode p = null;
        while(queue.size() > 0){
            int level1 = 0;//下一层需要出队的个数
            while(count < level - 1){
                p = queue.poll();
                p.next = queue.peek();
                if(p.left != null){
                    queue.offer(p.left);
                    level1++;
                }
                if(p.right != null){
                    level1++;
                    queue.offer(p.right);
                }
                count++;
            }
            p = queue.poll();
            if(p.left != null){
                queue.offer(p.left);
                level1++;
            }
            if(p.right != null){
                level1++;
                queue.offer(p.right);
            }
            level = level1;
            count = 0;
        }
    }
}

Populating Next Right Pointers in Each Node II

标签：leetcode

原文地址：http://blog.csdn.net/havedream_one/article/details/42555293