Leetcode 116. Populating Next Right Pointers in Each Node

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本文介绍如何使用队列和递归方法填充完美二叉树中每个节点的下一个指针，使其指向其右侧相邻节点。同时提供两种实现方式，并通过示例展示最终效果。

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;
        }
    }