117. Populating Next Right Pointers in Each Node II

题目：

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

 

 

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  Tree Depth-first Search  

链接： http://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/

题解：

一样是DFS。跟上题不同在于，给定输入不是完全二叉树了，所以要加入一些条件来判断是否存在一个合理的next值。并且对左节点和右节点的有效性也要验证。最后要先递归连接右节点，再connect左节点。

Time Complexity - O(n)， Space Complexity - O(1)。

public class Solution {

    public void connect(TreeLinkNode root) {

        if(root == null)

            return;

        TreeLinkNode node = root.next;

        while(node != null){

            if(node.left != null){

                node = node.left;

                break;

            } else if(node.right != null){

                node = node.right;

                break;

            } 

            node = node.next;

        }

        

        if(root.right != null){

            root.right.next = node;

            if(root.left != null)

                root.left.next = root.right;

        } else {

            if(root.left != null)

                root.left.next = node;

        }

        

        connect(root.right);

        connect(root.left);

    }

}

 

测试：

Reference:

http://www.cnblogs.com/springfor/p/3889327.html