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Populating Next Right Pointers in Each Node II <leetcode>

2024-07-21 19:18:50   220人阅读

Follow up for problem "Populating Next Right Pointers in Each Node".

What if the given tree could be any binary tree? Would your previous solution still work?

Note:

  • You may only use constant extra space.

 

For example,
Given the following binary tree,

         1       /        2    3     / \        4   5    7

 

After calling your function, the tree should look like:

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

算法：该题和第一个题的变种，不过揭发是一样的，首先也是只要考虑当前节点，然后递归就行了，要点有二：1、当该节点左子树或右子树的右边的第一个节点   2、递归时要先算右子树，因为左子树的处理递归时会用到右边已经形成的结果，代码如下：

 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(NULL==root)  return;13         if(NULL!=root->left)14         {15              if(NULL!=root->right)16              {17                  root->left->next=root->right;18              }19              else  root->left->next=find(root->next);20         }21         if(NULL!=root->right)22         {23             root->right->next=find(root->next);24         }25         26         27         connect(root->right);28         connect(root->left);29     }30     31     TreeLinkNode* find(TreeLinkNode *temp)32     {33         if(NULL==temp)  return NULL;34         while(NULL!=temp)35         {36             if(NULL!=temp->left)  return temp->left;37             else if(NULL!=temp->right)  return temp->right;38             else temp=temp->next;39         }40         return NULL;41     }42 };

 

Populating Next Right Pointers in Each Node II <leetcode>


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