path sum2

Change-Id: I53f9100815f68d914630525e4dd897f3b9715757

Author: applewjg

PathSum2.java

@@ -0,0 +1,57 @@
+/*
+ Author:     King, [email protected]
+ Date:       Oct 7, 2014
+ Problem:    Path Sum 2
+ Difficulty: easy
+ Source:     https://oj.leetcode.com/problems/path-sum-ii/
+ Notes:
+ Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.
+
+ For example:
+ Given the below binary tree and sum = 22,
+              5
+             / \
+            4   8
+           /   / \
+          11  13  4
+         /  \    / \
+        7    2  5   1
+ return
+ [
+   [5,4,11,2],
+   [5,8,4,5]
+ ]
+ 
+ Solution: DFS. 
+ */
+
+ /**
+ * Definition for binary tree
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode(int x) { val = x; }
+ * }
+ */
+public class Solution {
+    public List<List<Integer>> pathSum(TreeNode root, int sum) {
+        List<List<Integer>> res = new ArrayList<List<Integer>>();
+        ArrayList<Integer> path = new ArrayList<Integer>();
+        pathSumRe(root, sum, res, path);
+        return res; 
+    }
+    public void pathSumRe(TreeNode root, int sum, List<List<Integer>> res, ArrayList<Integer> path)
+    {
+        if (root == null) return;
+        path.add(root.val);
+        if (root.left == null && root.right == null && root.val == sum){
+            ArrayList<Integer> tmp = new ArrayList<Integer>(path);
+            res.add(tmp);
+        }
+        pathSumRe(root.left, sum - root.val, res, path);
+        pathSumRe(root.right, sum - root.val, res, path);
+        path.remove(path.size()-1);
+    }
+}
+