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 

方法一：dfs

时间复杂度：O(N2)

空间复杂度：O(N)

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @param {number} targetSum
 * @return {number}
 */

var psthSum = function(root, targetSum) {
    if (root === null) {
        return 0;
    }
    let ret = rootSum(root, targetSum);
    ret += pathSum(root.left, targetSum);
    ret += pathSum(root.right, targetSum);
    return ret;
}

const rootSum = (root, targetSum) => {
        let ret = 0;
        if (root === null) {
            return 0;
        }
        const val = root.val;
        if (val === targetSum) {
            ret++;
        }
        ret += rootSum(root.left, targetSum - val);
        ret += rootSum(root.right, targetSum - val);
        return ret;
    }

方法二：前缀和

时间复杂度：O(N)

空间复杂度：O(N)

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @param {number} targetSum
 * @return {number}
 */

var psthSum = function(root, targetSum) {
    const prefix = new Map();
    prefix.set(0, 1);
    return dfs(root, prefix, 0, targetSum);
    const dfs = (root, prefix, curr, targetSum);
}
const dfs = (root, prefix, curr, targetSum) => {
        if (root === null) {
            return 0;
        }
        let ret = 0;
        curr += root.val;
        ret = prefix.get(curr - targetSum) || 0;
        prefix.set(curr, (prefix.get(curr) || 0) + 1);
        ret += dfs(root.left, prefix, curr, targetSum);
        ret += dfs(root.right, prefix, curr, targetSum);
        prefix.set(curr, (prefix.get(curr) || 0) - 1);
        return ret;
    }