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  • [LeetCode] 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
    
    class Solution {
    public:
        void connect(TreeLinkNode *root) {
            if(root==NULL)
                return;
            
            root->next = NULL;
            fun(root);
        }
    private:
        TreeLinkNode *FindPleft(TreeLinkNode *&parent){
            TreeLinkNode *pleft =NULL;
            while(parent!=NULL){
                if(parent->left!=NULL){
                    pleft = parent->left;
                    break;
                }else if(parent->left==NULL && parent->right!=NULL){
                   pleft = parent->right;
                   break;
                }else{
                   parent = parent->next;
                   continue;
                }
            }//end while
            return pleft;
        }
        void fun(TreeLinkNode *parent){
            if(parent==NULL)
                return ;
            TreeLinkNode *pleft = FindPleft(parent);
            TreeLinkNode *pl;
            TreeLinkNode *nextLevelParent = pleft;
            while(parent!=NULL){
              if(pleft==parent->left &&parent->right!=NULL){
               pleft->next = parent->right;
               pleft = pleft->next;
            
              }else if((pleft!=NULL && pleft==parent->left && parent->right==NULL)||pleft==parent->right){
                  parent = parent->next;
                  if(parent==NULL||FindPleft(parent)==NULL){
                      pleft->next = NULL;
                      parent = nextLevelParent;
                      pl = FindPleft(parent);
                      pleft = pl;
                      nextLevelParent = pleft;
                  }else{
                      pl = FindPleft(parent);
                      pleft->next = pl;
                      pleft = pl;
                  }
              }//end if
                     
            }//end while
        }
    };
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  • 原文地址:https://www.cnblogs.com/Xylophone/p/3889542.html
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