leetcode 116 Populating Next Right Pointers in Each Node ----- java

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

改变树的结构。

第一种用队列，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() ){

            TreeLinkNode node = (TreeLinkNode) queue.poll();

            if( node.left != null){
                node.left.next = node.right;
                if( node.next != null )
                    node.right.next = node.next.left;
                queue.add(node.left);
                queue.add(node.right);
            }
        }

    }
}

不使用队列就会达到最快。

/**
 * 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 ;
        TreeLinkNode node1 = root,node2 = root;
        while( node2.left != null ){

            node1 = node2.left;
            while( node2.next != null ){
                node2.left.next = node2.right;
                node2.right.next = node2.next.left;
                node2 = node2.next;
            }
            node2.left.next = node2.right;
            node2 = node1;

        }

    }
}