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:

https://www.jiuzhang.com/solution/path-sum-iii/

https://leetcode.com/problems/path-sum-iii/discuss/91878/17-ms-O(n)-java-Prefix-sum-method

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;
    }

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