Leetcode 437. Path Sum III 路径和3 解题报告

1 解题思想

这道题就是给了一个二叉树和一个目标和sum
找出所有路径，这个路径的和等于sum，只允许从父节点到子节点的路线

所以方法么，也就是最基本的dfs，不多说，对了可以看看之前相关的题目：
Leetcode 64. Minimum Path Sum 最小路径和 解题报告
Leetcode 112. Path Sum 路径和 解题报告
Leetcode 113. Path Sum II 路径和2 解题报告

2 原题

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

3 AC解

/** * 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 dfs(root, sum)+pathSum(root.left, sum)+pathSum(root.right, sum);
    }

    private int dfs(TreeNode root, int sum){
        int res = 0;
        if(root == null)
            return res;
        if(sum == root.val)
            res++;
        res+=dfs(root.left,sum - root.val);
        res+=dfs(root.right,sum - root.val);
        return res;
    }
}