Leetcode 116. Populating Next Right Pointers in Each Node

Question

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

Code

队列方法

public void connect(TreeLinkNode root) {
   if (root == null) {
            return;
        }

        Queue<TreeLinkNode> q = new LinkedList<TreeLinkNode>();
        q.offer(root);

        while (!q.isEmpty()) {
            int size = q.size();
            for (int i = 0; i < size; i++) {
                TreeLinkNode cur = q.poll();
                cur.next = (i == size - 1) ? null : q.peek();
                if (cur.left != null) {
                    q.offer(cur.left);
                    q.offer(cur.right);
                }
            }
        }
    }

方法二递归方式

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

      rec(root);

    }

    public void rec(TreeLinkNode root) {
        if (root == null) {
            return;
        }

        if (root.left != null) {
            root.left.next = root.right;
        }

        if (root.right != null) {
            root.right.next = root.next == null ? null : root.next.left;
        }

        rec(root.left);
        rec(root.right);
    }

good way

 public void connect(TreeLinkNode root) {
      if (root == null) {
            return;
        }

        TreeLinkNode leftEnd = root;
        while (leftEnd != null && leftEnd.left != null) {
            TreeLinkNode cur = leftEnd;
            while (cur != null) {
                cur.left.next = cur.right;
                cur.right.next = cur.next == null ? null : cur.next.left;

                cur = cur.next;
            }

            leftEnd = leftEnd.left;
        }
    }