Populating Next Right Pointers in Each Node

已于 2022-03-27 13:10:38 修改 · 406 阅读

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于 2014-12-07 04:27:51 首次发布

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本文详细介绍了如何使用常数额外空间来填充完美二叉树中每个节点的 next 指针，使其指向其右侧的节点。通过遍历树并连接相邻节点，实现树结构的优化。

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 toNULL.

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

思路: 就是主要连接两个，一个是root.left != NULL, root.right != null ，左边节点往右，右边节点往右，然后root往后挪动，当前层连接完了，root变成firstNode.left;

class Solution {
    public Node connect(Node root) {
        if(root == null) {
            return null;
        }
        Node head = root;
        while(root != null) {
            Node firstNode = root;
            while(root != null) {
                if(root.left != null) {
                    root.left.next = root.right;
                }
                if(root.right != null && root.next != null) {
                    root.right.next = root.next.left;
                }
                root = root.next;
            }
            root = firstNode.left;
        }
        return head;
    }
}