题目描述:

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:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
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

这道题用很直接的遍历方法就可以AC, 所以才会是Easy. 我还想了很久有没有O(n)或者O(nlogn)的办法…

对于每一棵子树, 都计算从根节点开始的路径和有没有等于sum的, 这一步用DFS遍历所有路径. 然后递归地处理左子树和右子树.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
    int ans = 0;
public:
    int pathSum(TreeNode* root, int sum) {
        if(!root) return 0;
        DFS(root, sum);
        pathSum(root->left, sum);
        pathSum(root->right, sum);
        return ans;
    }
    
    void DFS(TreeNode *node, int target){
        if(!node) return;
        target -= node->val;
        if(0 == target) ans++;
        DFS(node->left, target);
        DFS(node->right, target);
        target += node->val;
    }
};