【leetcode hot 100 437】路径总和Ⅲ

解法一：（深度优先搜索）穷举所有的可能 =》访问每一个节点 node，检测以 node 为起始节点且向下延深的路径有多少种。递归遍历每一个节点的所有可能的路径，然后将这些路径数目加起来即为返回结果。

/**
 * 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) {
        if(root==null){
            return 0;
        }

        int res = rootSum(root, targetSum);
        res += pathSum(root.left, targetSum);
        res += pathSum(root.right, targetSum);
        return res;
    }

    public int rootSum(TreeNode root, int targetSum){
        if(root==null){
            return 0;
        }
        int res = 0;
        if(root.val==targetSum){
            res++;
        }
        res += rootSum(root.left, targetSum-root.val);
        res += rootSum(root.right, targetSum-root.val);
        return res;
    }
}

注意：

• public int pathSum(TreeNode root, int targetSum)：根节点+左节点+右节点分开单独遍历每一个节点的所有可能的路径；public int rootSum(TreeNode root, int targetSum)遍历根节点所有可能的路径。
• 将pathSum(TreeNode root, int targetSum)和rootSum(TreeNode root, int targetSum)的int改为long可以解决下述问题。