222. Count Complete Tree Nodes

Given a complete binary tree, count the number of nodes.

Definition of a complete binary tree from Wikipedia:
In a complete binary tree every level, except possibly the last, was completely filled, and all nodes As far left as possible. It can has between 1 and 2h nodes inclusive at the last level H.

The direct traversal will time out and the code is as follows:

1 classSolution {2  Public:3     intCountnodes (treenode*root) {4         intCount =0;5Queue<treenode *>temp;6         if(Root = =NULL)7         {8             return 0;9         }Ten Temp.push (root); One          while(!temp.empty ()) A         { -TreeNode * Node =Temp.front (); -count++; the Temp.pop (); -              -             if(Node->left! =NULL) -             { +Temp.push (node->Left ); -             } +             if(Node->right! =NULL) A             { atTemp.push (node->Right ); -             } -         } -         returncount; -         } -};

This problem to find the total number of nodes of the binary tree, so for a node, if the number of left subtree traversal and the number of right sub-tree traversal is equal, it is full of two fork tree, then the formula is 2^h-1, if not equal is not full two fork tree, recursive calculation Saozi right subtree plus the current node can be. The code is as follows:

1 classSolution {2  Public:3     intCountnodes (treenode*root) {4             intLeftnum =0;5             intRightnum =0;6TreeNode * ptr =Root;7         8             if(Root = =NULL)9             {Ten                 return 0; One             } A          -              while(1) -             { thePTR = ptr->Left ; -                 if(PTR! =NULL) -                 { -++Leftnum; +                 } -                 Else +                 { A                      Break; at                 } -             } -PTR =Root; -              while(1) -             { -PTR = ptr->Right ; in                 if(PTR! =NULL) -                 { to++Rightnum; +                 } -                 Else the                 { *                      Break; $                 }Panax Notoginseng             } -             if(Leftnum = =0&& Rightnum = =0) the             { +                 return 1; A             } the             if(Leftnum = =rightnum) +             { -                 returnPow2, Leftnum +1) -1; $             } $             return 1+ countnodes (root->left) + countnodes (root->Right ); -         } -};

222. Count Complete Tree Nodes