Question

link

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

Stats

Frequency 3
Difficulty 3
Adjusted Difficulty 2
Time to use --------

Ratings/Color = 1(white) 2(lime) 3(yellow) 4/5(red)

Analysis

Although this question is not hard, it needs a bit thinking.

The important point is, how to effectively make use of the ‘next link’ generated before, to help us solve this problem.

Solution

My solution is actually very good. It pass current node and next node into a method, and then generate links for current node’s children.

There is a even better solution, which directly make use of the ‘next link’ generated already. In fact, it’s same as my solution, except it uses one less variable. Great idea it is!

Code

First, my solution

public void connect(TreeLinkNode root) {
    link(root, null);
}

private void link(TreeLinkNode node, TreeLinkNode rr){
    if (node == null || node.left == null) return;
    node.left.next = node.right;
    if (rr == null) node.right.next = null;
    else node.right.next = rr.left;

    link(node.left, node.left.next);
    link(node.right, node.right.next);
}

Second, better solution

public void connect(TreeLinkNode root) {
    if (root == null || root.left == null || root.right == null)
        return;
    root.left.next = root.right;
    root.right.next = root.next == null ? null : root.next.left;
    connect(root.left);
    connect(root.right);
}