[LeetCode]Populating Next Right Pointers in Each Node

最新推荐文章于 2021-09-24 12:05:40 发布

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最新推荐文章于 2021-09-24 12:05:40 发布

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原文链接：http://www.cnblogs.com/Sean-le/p/4783111.html

本文介绍了一种在二叉树中填充每个节点的next指针的方法，使得每个节点的next指针指向其右侧的下一个节点。提供了两种实现方式：一种是递归方式，另一种是一次性遍历一行的方式。

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

递归实现，注意到上一层在下一层之前完成next的指向。

 1 /**
 2  * Definition for binary tree with next pointer.
 3  * struct TreeLinkNode {
 4  *  int val;
 5  *  TreeLinkNode *left, *right, *next;
 6  *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     void connect(TreeLinkNode *root) {
12         if(!root || !root->left) return;
13         root->left->next = root->right;
14         if(root->next)
15         {
16             root->right->next = root->next->left;
17         }
18         connect(root->left);
19         connect(root->right);
20     }
21 };

还有一种方法就是，每次一次性遍历一整行。这样只要tree的最左边节点就可以了。

 1 /**
 2  * Definition for binary tree with next pointer.
 3  * struct TreeLinkNode {
 4  *  int val;
 5  *  TreeLinkNode *left, *right, *next;
 6  *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     void connect(TreeLinkNode *root) {
12         if(!root || !root->left) return;
13         TreeLinkNode* p = root;
14         while(p)
15         {
16             p->left->next = p->right;
17             if(p->next)
18             {
19                 p->right->next = p->next->left;
20             }
21             p = p->next;
22         }
23         connect(root->left);
24     }
25 };

转载于:https://www.cnblogs.com/Sean-le/p/4783111.html