Populating Next Right Pointers in Each Node

最新推荐文章于 2022-02-28 18:16:11 发布

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本文探讨如何使用递归方法解决二叉树节点连接问题，通过先根、右、左节点的顺序遍历来简化操作，最终实现每个节点指向其右侧相邻节点的功能。

题目：

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

这个题目采用递归的思路，如果上一层和右边的已经处理好next指针，那么处理当前节点就比较容易了。因此在进行递归的时候就应该是采用先根、然后右、然后左节点的顺序进行递归遍历。这个思路在Flatten Binary Tree to Linked List My Submissions Question Solution这道题目中也有体现，也是从右节点开始处理。

/**
 * 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(null == root) return ;
        TreeLinkNode rootnext = root.next;
        TreeLinkNode next = null;
        while (rootnext != null && next == null){
            if(rootnext.left != null){
                next = rootnext.left;
            }else{
                next = rootnext.right;
            }
            rootnext = rootnext.next;
        }
        if(root.left != null){
            if(root.right != null){
                root.left.next = root.right;
            }else{
                root.left.next = next;
            }
        }
        if(root.right != null){
            root.right.next = next;
        }
        connect(root.right);
        connect(root.left);
    }
}