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

这里分别记录递归解法和hashmap解法：
递归解法，思路清晰，遍历各个子树，查找各个子路径，寻找sum=target的子路径，时间复杂度O(n^2)，代码如下：

/**
* 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;
}
return findSub(root, sum) + pathSum(root.left, sum) + pathSum(root.right, sum);
}
public int findSub(TreeNode root, int sum) {
int res = 0;
if (root == null) {
return 0;
}
if (sum == root.val) {
res ++;
}
res += findSub(root.left, sum - root.val);
res += findSub(root.right, sum - root.val);
return res;
}
}Hashmap解法，时间复杂度O(n)，代码如下：

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;
}