LeetCode 116. Populating Next Right Pointers in Each Node（右邻）

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本文介绍了一种在完美二叉树中填充每个节点的next指针的方法，使其指向其右侧相邻节点。提供了两种实现方式：递归方法及优化后的递归调用次数的方法，并附带详细的代码示例。

原题网址：https://leetcode.com/problems/populating-next-right-pointers-in-each-node/

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,

After calling your function, the tree should look like:

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \  / \
    4->5->6->7 -> NULL

方法一：递归。

/**
 * Definition for binary tree with next pointer.
 * public class TreeLinkNode {
 *     int val;
 *     TreeLinkNode left, right, next;
 *     TreeLinkNode(int x) { val = x; }
 * }
 */
public class Solution {
    private void connect(TreeLinkNode left, TreeLinkNode right) {
        if (left == null || right == null) return;
        left.next = right;
        connect(left.left, left.right);
        connect(right.left, right.right);
        connect(left.right, right.left);
    }
    public void connect(TreeLinkNode root) {
        if (root == null) return;
        connect(root.left, root.right);
    }
}

方法二：优化递归调用次数。

/**
 * 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 || root.right == null) return;
        connect(root.left);
        connect(root.right);
        TreeLinkNode node1 = root.left;
        TreeLinkNode node2 = root.right;
        node1.next = node2;
        while (node1.right != null) {
            node1 = node1.right;
            node2 = node2.left;
            node1.next = node2;
        }
    }
}