116. Populating Next Right Pointers in Each Node

Difficulty: Medium
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

当root的左子树为空的时候，可以直接返回； 当root的左子树不为空时，左子树的next为root的右子树；但右子树要分情况讨论：

• 当root.next为空时，root.right的next也为空
• 当root.next不为空时，root.right的next为root.next.left.

然后递归connect根节点的左右子树。

/**
 * 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 || root.left == null){
            return;
        }
        if (root.left != null){
            root.left.next = root.right;
        } 
        if (root.next != null){
            root.right.next = root.next.left;
        }
        connect(root.left);
        connect(root.right);
    }
}