437. 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

public class Solution {
    public int pathSum(TreeNode root, int sum) {
        if (root == null) return 0;
        return pathSumFrom(root, sum) + pathSum(root.left, sum) + pathSum(root.right, sum);
    }
    
    private int pathSumFrom(TreeNode node, int sum) {
        if (node == null) return 0;
        return (node.val == sum ? 1 : 0) 
            + pathSumFrom(node.left, sum - node.val) + pathSumFrom(node.right, sum - node.val);
    }
}

巧啊，给pathsum自己call自己，这样相当于在root.left, root.rigtht处从上到下从root到val，然后pathsumfrom就正常从root到null即可

 

public int pathSum(TreeNode root, int sum) {
        if (root == null) {
            return 0;
        }
        Map<Integer, Integer> map = new HashMap<>();
        map.put(0, 1);
        return findPathSum(root, 0, sum, map);  
    }
    private int findPathSum(TreeNode curr, int sum, int target, Map<Integer, Integer> map) {
        if (curr == null) {
            return 0;
        }
        // update the prefix sum by adding the current val
        sum += curr.val;
        // get the number of valid path, ended by the current node
        int numPathToCurr = map.getOrDefault(sum-target, 0); 
        // update the map with the current sum, so the map is good to be passed to the next recursion
        map.put(sum, map.getOrDefault(sum, 0) + 1);
        // add the 3 parts discussed in 8. together
        int res = numPathToCurr + findPathSum(curr.left, sum, target, map)
                                               + findPathSum(curr.right, sum, target, map);
       // restore the map, as the recursion goes from the bottom to the top
        map.put(sum, map.get(sum) - 1);
        return res;
    }

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

用prefix sum，prefix sum是从root加到当前node的sum，那么我们可以想到设x，然后prefix sum = x + target的话，x =  prefixsum - target，其中x是从root到当前node中的一点，这样的话从这个x点到cur node的和就是target了。

然后又因为x在map中，所以x的频率就是root到当前node能形成target sum的数量。随后更新sum的频率，继续dfs，这个点dfs完后要进行剪枝（因为这个点的 情况考虑完了，所以不要再用这个点了）

return res