leetcode 117 Populating Next Right Pointers in Each Node II ----- java

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

这道题和上一道题的区别在于，上一道的树是满二叉树，这一个并不是。

还是先使用队列做了一次，ac但是速度并不是很快。

/**
 * 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 queue = new LinkedList<TreeLinkNode>();

        queue.add(root);

        while( !queue.isEmpty() ){

            int size = queue.size();
            TreeLinkNode node1 = (TreeLinkNode) queue.poll();
            if( node1.left != null )
                queue.add(node1.left);
            if( node1.right != null)
                queue.add(node1.right);
            if( size == 1)
                continue;
            TreeLinkNode node2 = (TreeLinkNode) queue.poll();
            if( node2.left != null )
                queue.add(node2.left);
            if( node2.right != null)
                queue.add(node2.right);
            for( int i = 2;i<size;i++){
                node1.next = node2;
                node1 = node2;
                node2 = ( TreeLinkNode ) queue.poll();
                if( node2.left != null )
                    queue.add(node2.left);
                if( node2.right != null)
                    queue.add(node2.right);
            }
            node1.next = node2;
        }

    }
}

但是题目中要求是常数空间。

所以还需要修改。

记录上一行的开始和下一行的开始，然后依次改变next。

/**
 * 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 ;
        TreeLinkNode low = null ;//指的是下面一行的第一个结点
        TreeLinkNode up = root;
        if( root.left != null )
            low = root.left;
        else if( root.right != null )
            low = root.right;
        while( low != null ){
            TreeLinkNode start = low;
            TreeLinkNode upStart = up;
            helper(start,upStart);
            while( low != null ){
                if( low.left != null ){
                    TreeLinkNode node = low.left;
                    up = low;
                    low = node;
                    break;
                }
                if( low.right != null ){
                    TreeLinkNode node = low.right;
                    up = low;
                    low = node;
                    break;
                }
                low = low.next;
            }

        }
    }

    public void helper(TreeLinkNode start,TreeLinkNode upStart){

        if( upStart.left != null){
            if( upStart.right != null){
                start.next = upStart.right;
                start = start.next;
            }
        }
        upStart = upStart.next;
        while( upStart != null ){

            if( upStart.left != null  ){
                start.next = upStart.left;
                start = start.next;
                if( upStart.right != null ){
                    start.next = upStart.right;
                    start = start.next;
                }
            }else if( upStart.right != null ){
                start.next = upStart.right;
                start = start.next;
            }
            upStart = upStart.next;
        }
    }
}