add solution of problem 113: path sum II

Author: ppulse

PathSumII113/README.md

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+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]
+
+]
+
+```

PathSumII113/Solution.java

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+/**
+ * Definition for a binary tree node.
+ * 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>> result = new ArrayList<List<Integer>>();
+        
+        if (root == null)
+            return result;
+        
+        Stack<Integer> path = new Stack<Integer>();
+        preOrder(root, path, result, 0, sum);
+        return result;
+    }
+    
+    private final void preOrder(TreeNode node, Stack<Integer> path, List<List<Integer>> result, int currSum, int expectedSum) {
+        path.push(node.val);
+        currSum += node.val;
+        
+        if (node.left == null && node.right == null && currSum == expectedSum) {
+            if (currSum == expectedSum) {
+                List<Integer> pathList = new ArrayList<Integer>();
+                for (int val : path) {
+                    pathList.add(val);
+                }
+                
+                result.add(pathList);
+            }
+        }
+        
+        if (node.left != null) {
+            preOrder(node.left, path, result, currSum, expectedSum);
+        }
+        
+        if (node.right != null) {
+            preOrder(node.right, path, result, currSum, expectedSum);
+        }
+        
+        path.pop();
+    }
+}