Sum Root to Leaf Numbers

Sum Root to Leaf Numbers 

Given a binary tree containing digits from only 0-9 , each root-to-leaf path could represent a number.

An example is the Root-to-leaf path 1->2->3 which represents the number 123 .

Find The total sum of all root-to-leaf numbers.

For example,

    1   /   2   3

The Root-to-leaf path 1->2 represents the number 12 .
The Root-to-leaf path 1->3 represents the number 13 .

Return the sum = + = 25 .

This problem, I was on the basis of sequential traversal, added a similar stack of the list, if the non-leaf node will be "into the stack", if it is a leaf node "stack" until node's father nodes appear, scan "bottom" to "stack top" all elements of the number, calculate sum. Until the end of the pre-order traversal

1 Importjava.util.ArrayList;2 Importjava.util.List;3 4 5 6 7    classTreeNode {8       intVal;9 TreeNode left;Ten TreeNode right; OneTreeNode (intX) {val =x;} A   } -  -  Public classSolution { the  -list<treenode> list =NewArraylist<treenode>(); -      -     intsum = 0; +      -      Public intsumnumbers (TreeNode root) { +         if(NULL= = root)//empty tree returns 0 A             return0; at         Else if(NULL= = Root.left &&NULL= = Root.right) {//only the root node returns the value of the root node -             returnRoot.val; -}Else{ -Preorder (root);//there is at least one sub-tree - //List.add (root); //root node into the stack -         } in          -         returnsum; to     } +      -     /** the * Pre-sequence traversal tree *      * @paramRoot $      */Panax Notoginseng      Public voidpreorder (TreeNode root) { -         if(Root = =NULL) the             return; +         if(NULL= = Root.left && Root.right = =NULL){//leaf node A              while(List.size ()! = 0 && Root! = List.get (List.size ()-1). Left && root! = List.get (List.size ()-1). right) { theList.remove (List.size ()-1); +             } -             inttemp = 0; $              for(inti = 0; I < list.size (); i++) {//calculates a number $temp = temp * 10 +List.get (i). val; -             } -Temp = temp * + root.val;//leaf node plus go the             //System.out.println ("temp =" + temp); -sum + = temp;//Calculation andWuyi             if(Root = = List.get (List.size ()-1). right) {//if the leaf node is the right sub-tree, the stack theList.remove (List.size ()-1); -             } Wu         } -         Else{//Non-leaf node About              while(List.size ()! = 0 && Root! = List.get (List.size ()-1). Left && root! = List.get (List.size ()-1). right) { $List.remove (List.size ()-1); -}//root should be the same as the top element of the stack - List.add (root); - preorder (root.left); A preorder (root.right); +         } the     } -}

The topic is to use DFS, a while Baidu a bit ~

Look at the inside of discuss, this is very beautiful written

 Public classSolution { Public intsumnumbers (TreeNode root) {returnPreorder (Root, 0); }         Public intPreorder (TreeNode root,intval) {        if(Root = =NULL)            return0; Val= val * 10 +Root.val; if(NULL= = Root.left &&NULL==root.right)returnVal; Else            returnPreorder (Root.left, Val) +Preorder (Root.right, Val); }}

Sum Root to Leaf Numbers