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 , nodes and the values are in the range -,, to ,,.

Example:

root = [,,-,,,null,,,-,null,], sum = 

     /  \
       -
   / \    \

 / \   \
  -   

Return . The paths that sum to  are:

.   ->
.   ->  ->
. - -> 

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

My Solution

 /**
  * Definition for a binary tree node.
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 public class Solution {
     public int pathSum(TreeNode root, int sum) {
         if (root == null) return 0;
         ArrayList<Integer> res = new ArrayList<Integer>();
         res.add(0);
         HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
         map.put(0, 1);
         helper(root, sum, 0, res, map);
         return res.get(0);
     }

     public void helper(TreeNode cur, int target, int pathSum, ArrayList<Integer> res, HashMap<Integer, Integer> map) {
         if (map.containsKey(pathSum+cur.val-target)) {
             res.set(0, res.get(0) + map.get(pathSum+cur.val-target));
         }
         if (!map.containsKey(pathSum+cur.val)) {
             map.put(pathSum+cur.val, 1);
         }
         else map.put(pathSum+cur.val, map.get(pathSum+cur.val)+1);
         if (cur.left != null) helper(cur.left, target, pathSum+cur.val, res, map);
         if (cur.right != null) helper(cur.right, target, pathSum+cur.val, res, map);
         map.put(pathSum+cur.val, map.get(pathSum+cur.val)-1);
     }
 }

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.

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