【leetcode】Populating Next Right Pointers in Each Node II

Populating Next Right Pointers in Each Node II

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

 

与I的思路类似

left->next=father->right

right->next=father->next->left;

 

只是当我们没有找到时，father=father->next;

需要注意，先搜索右子树，再搜索左子树

 

 

 /**
  * Definition for binary tree with next pointer.
  * struct TreeLinkNode {
  *  int val;
  *  TreeLinkNode *left, *right, *next;
  *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
  * };
  */
 class Solution {
 public:
     void connect(TreeLinkNode *root) {

         if(root==NULL)
         {
             return;
         }

         if(root->left!=NULL)
         {
             if(root->right!=NULL)
             {
                 root->left->next=root->right;
             }
             else
             {
                 TreeLinkNode *tmp=root->next;
                 root->left->next=NULL;
                 while(tmp!=NULL)
                 {
                     if(tmp->left!=NULL)
                     {
                         root->left->next=tmp->left;
                         break;
                     }

                     if(tmp->right!=NULL)
                     {
                         root->left->next=tmp->right;
                         break;
                     }
                     tmp=tmp->next;
                 }
             }
         }

         if(root->right!=NULL)
         {
             TreeLinkNode *tmp=root->next;
             root->right->next=NULL;
             while(tmp!=NULL)
             {
                 if(tmp->left!=NULL)
                 {
                     root->right->next=tmp->left;
                     break;
                 }

                 if(tmp->right!=NULL)
                 {
                     root->right->next=tmp->right;
                     break;
                 }
                 tmp=tmp->next;
             }
         }

         connect(root->right);
         connect(root->left);

     }
 };