257. Binary Tree Paths

Given a binary tree, return all root-to-leaf paths.

For example, given the following binary tree:

   1/   2     3   5

All root-to-leaf paths is:

["1->2->5", "1->3"]

Idea: DFS, search to leaf stop, save to Res. I used the StringBuilder, in the back of the time how to return to the previous state did not think out. Reference to a method in the discussion, SetLength. Mactan, almost, look at the answer is much more comfortable.

/*** Definition for a binary tree node. * public class TreeNode {* int val; * TreeNode left; * TreeNode rig Ht * TreeNode (int x) {val = x;} }*/ Public classSolution { PublicList<string>binarytreepaths (TreeNode root) {List<String> res=NewArraylist<string>(); StringBuilder Save=NewStringBuilder ();        DPS (Root,res,save); returnRes; }     Public voidDPS (TreeNode root, list<string>Res, StringBuilder SB) {        if(root==NULL)        {            return; }        intLength=sb.length (); if(root.left==NULL&&root.right==NULL) {sb.append (root.val);            Res.add (Sb.tostring ());            Sb.setlength (length); return;        } sb.append (Root.val); Sb.append (",");        DPS (ROOT.LEFT,RES,SB);        DPS (ROOT.RIGHT,RES,SB);    Sb.setlength (length); }            }

More understanding of the point of backtracking, that is, the condition of return.

257. Binary Tree Paths