437. 路径总和 III

给定一个二叉树的根节点 root ，和一个整数 targetSum ，求该二叉树里节点值之和等于 targetSum 的路径的数目。

路径不需要从根节点开始，也不需要在叶子节点结束，但是路径方向必须是向下的（只能从父节点到子节点）。

示例 1：

输入：root = [10,5,-3,3,2,null,11,3,-2,null,1], targetSum = 8
输出：3
解释：和等于 8 的路径有 3 条，如图所示。

示例 2：

输入：root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
输出：3

提示:

二叉树的节点个数的范围是 [0,1000]
109 <= Node.val <= 109
1000 <= targetSum <= 1000

Solution 1

记录访问过节点的值，每次访问逆序求和得出targetSum
时间复杂度O(n*h)，空间复杂度O(n)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 * int val;
 * TreeNode left;
 * TreeNode right;
 * TreeNode() {}
 * TreeNode(int val) { this.val = val; }
 * TreeNode(int val, TreeNode left, TreeNode right) {
 * this.val = val;
 * this.left = left;
 * this.right = right;
 * }
 * }
 */
class Solution {
    public int pathSum(TreeNode root, int targetSum) {
        List<Integer> stack = new ArrayList<>();
        List<Integer> cnt = new LinkedList<>();
        cnt.add(0);
        traverse(root, targetSum, stack, cnt);
        return cnt.remove(0);
    }

    void traverse(TreeNode root, int targetSum, List<Integer> stack, List<Integer> cnt) {
        if (root == null) {
            return;
        }
        stack.add(root.val);
        long sumVal = 0;
        for (int i = stack.size() - 1; i >= 0; i--) {
            sumVal += stack.get(i);
            if (sumVal == targetSum) {
                cnt.add(1 + cnt.remove(0));
            }
        }
        traverse(root.left, targetSum, stack, cnt);
        traverse(root.right, targetSum, stack, cnt);
        stack.remove(stack.size() - 1);
    }
}

Solution 2

记录访问过节点的值的前缀和的次数，每次访问通过prevPrefixSum + targetSum == currPrefixSum求出prevPrefixSum的出现次数，也就是Solution 1中能求得几次targetSum
时间复杂度O(n)，空间复杂度O(n)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 * int val;
 * TreeNode left;
 * TreeNode right;
 * TreeNode() {}
 * TreeNode(int val) { this.val = val; }
 * TreeNode(int val, TreeNode left, TreeNode right) {
 * this.val = val;
 * this.left = left;
 * this.right = right;
 * }
 * }
 */
class Solution {
    public int pathSum(TreeNode root, int targetSum) {
        Map<Long, Integer> prefixSumMap = new HashMap<>();
        Map<Integer, Integer> cnt = new HashMap<>();
        prefixSumMap.put(0L, 1);
        cnt.put(null, 0);
        traverse(root, targetSum, 0, prefixSumMap, cnt);
        return cnt.get(null);
    }

    void traverse(TreeNode root, int targetSum, long prevPrefixSum, Map<Long, Integer> prefixSumMap,
            Map<Integer, Integer> cnt) {
        if (root == null) {
            return;
        }
        long currPrefixSum = prevPrefixSum + root.val;
        // prevPrefixSum + targetSum == currPrefixSum;
        int c = prefixSumMap.getOrDefault(currPrefixSum - targetSum, 0);
        cnt.put(null, c + cnt.get(null));
        prefixSumMap.put(currPrefixSum, 1 + prefixSumMap.getOrDefault(currPrefixSum, 0));
        traverse(root.left, targetSum, currPrefixSum, prefixSumMap, cnt);
        traverse(root.right, targetSum, currPrefixSum, prefixSumMap, cnt);
        prefixSumMap.put(currPrefixSum, -1 + prefixSumMap.get(currPrefixSum));
    }
}