Leetcode: Path Sum III

You are given a binary tree in which each node contains an integer value.

Find the number of paths that sum to a given value.

The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).

The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.

Example:

root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8

      10
     /  \
    5   -3
   / \    \
  3   2   11
 / \   \
3  -2   1

Return 3. The paths that sum to 8 are:

1.  5 -> 3
2.  5 -> 2 -> 1
3. -3 -> 11

Add the prefix sum to the hashMap, and check along path if hashMap.contains(pathSum+cur.val-target);

My Solution

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public int pathSum(TreeNode root, int sum) {
12         if (root == null) return 0;
13         ArrayList<Integer> res = new ArrayList<Integer>();
14         res.add(0);
15         HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
16         map.put(0, 1);
17         helper(root, sum, 0, res, map);
18         return res.get(0);
19     }
20     
21     public void helper(TreeNode cur, int target, int pathSum, ArrayList<Integer> res, HashMap<Integer, Integer> map) {
22         if (map.containsKey(pathSum+cur.val-target)) {
23             res.set(0, res.get(0) + map.get(pathSum+cur.val-target));
24         }
25         if (!map.containsKey(pathSum+cur.val)) {
26             map.put(pathSum+cur.val, 1);
27         }
28         else map.put(pathSum+cur.val, map.get(pathSum+cur.val)+1);
29         if (cur.left != null) helper(cur.left, target, pathSum+cur.val, res, map);
30         if (cur.right != null) helper(cur.right, target, pathSum+cur.val, res, map);
31         map.put(pathSum+cur.val, map.get(pathSum+cur.val)-1);
32     }
33 }

一个更简洁的solution: using HashMap to store ( key : the prefix sum, value : how many ways get to this prefix sum) , and whenever reach a node, we check if prefix sum - target exists in hashmap or not, if it does, we added up the ways of prefix sum - target into res.

 1     public int pathSum(TreeNode root, int sum) {
 2         Map<Integer, Integer> map = new HashMap<>();
 3         map.put(0, 1);  //Default sum = 0 has one count
 4         return backtrack(root, 0, sum, map); 
 5     }
 6     //BackTrack one pass
 7     public int backtrack(TreeNode root, int sum, int target, Map<Integer, Integer> map){
 8         if(root == null)
 9             return 0;
10         sum += root.val;
11         int res = map.getOrDefault(sum - target, 0);    //See if there is a subarray sum equals to target
12         map.put(sum, map.getOrDefault(sum, 0)+1);
13         //Extend to left and right child
14         res += backtrack(root.left, sum, target, map) + backtrack(root.right, sum, target, map);
15         map.put(sum, map.get(sum)-1);   //Remove the current node so it wont affect other path
16         return res;
17     }

posted @ 2016-12-07 12:17 neverlandly 阅读(599) 评论(0) 编辑收藏举报

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