Lintcode:binary Tree Path Sum

Description:

Given a binary tree, find all paths this sum of the nodes in the path equals to a Given number target .

A valid path is from root node to any of the leaf nodes.

Example:

Given a binary tree, and target = 5 :

     1    /    2   4  /  2   3

Return

[  [1, 2, 2],  [1, 4]]

/*** Definition of TreeNode: * public class TreeNode {* public int val; * Public TreeNode left, right; * PU Blic TreeNode (int val) {* This.val = val; * This.left = This.right = null; *} *}*/ Public classSolution {/**     * @paramroot the binary tree *@paramtarget an integer *@returnAll valid Paths*/    Privatelist<list<integer>> result =NewArraylist<>(); Privatearraylist<integer> Path =NewArraylist<integer>();  PublicList<list<integer>> binarytreepathsum (TreeNode root,inttarget) {        //Write Your code here//Handel Corner Cases        if(Root = =NULL) {            returnresult;        } path.add (Root.val); Helper (root, path, target-root.val); returnresult; }        Private voidHelper (TreeNode node, arraylist<integer> path,inttarget) {        if(Node.left = =NULL&& Node.right = =NULL) {            if(target = = 0) {Result.add (NewArraylist<integer>(path)); }            return; }        if(Node.left! =NULL) {path.add (node.left.val); Helper (node.left, path, Target-node.left.val); Path.remove (Path.size ()-1); }        if(Node.right! =NULL) {path.add (node.right.val); Helper (node.right, path, Target-node.right.val); Path.remove (Path.size ()-1); }    }}

View Code

With the divide and conquer method, the most important thing, the "to remove", the last value of T.left or Root.right, and also remember that if node.left = = null and Node.right = NULL, then it's time to return.

Lintcode:binary Tree Path Sum