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

Analysis

DFS
Traversing, two level of DFS
Similar to subarray sum:
具体操作:
用一个hashmap记录prefix sum和它的frequency (这道题要求所有路径,所以要记录frequency)
在dfs的过程中,维持一个current的变量,叠加本条路径的path sum
如果在某个点上current - target的差已经存在hashmap里了,就累计hashmap里的frequency
继续往左往右traverse下去
backtrack:
完成了一个点的traverse就把current的frequency减1.

Solution

DFS - Intuitive Implementation
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int pathSum(TreeNode root, int sum) {
if (root == null) {
return 0;
}
return pathHelper(root, sum) + pathSum(root.left, sum) + pathSum(root.right, sum);
}
int pathHelper(TreeNode root, int sum) {
if (root == null) return 0;
return (root.val == sum ? 1 : 0) +
pathHelper(root.left, sum - root.val) + pathHelper(root.right, sum - root.val);
}
}
(Better) Prefix Sum
public int pathSum(TreeNode root, int sum) {
HashMap<Integer, Integer> preSum = new HashMap();
preSum.put(0,1);
return helper(root, 0, sum, preSum);
}
public int helper(TreeNode root, int currSum, int target, HashMap<Integer, Integer> preSum) {
if (root == null) {
return 0;
}
currSum += root.val;
int res = preSum.getOrDefault(currSum - target, 0);
preSum.put(currSum, preSum.getOrDefault(currSum, 0) + 1);
res += helper(root.left, currSum, target, preSum) + helper(root.right, currSum, target, preSum);
preSum.put(currSum, preSum.get(currSum) - 1);
return res;
}