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[LeetCode] Path Sum III

时间:2017-07-24 23:22:52      阅读:295      评论:0      收藏:0      [点我收藏+]

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
题目要求一棵树中节点和为指定数值的路径数。
class Solution {
public:
    int pathSum(TreeNode* root, int sum) {
        if (root == nullptr)
            return 0;
        return dfs(root, sum) + pathSum(root->left, sum) + pathSum(root->right, sum);
    }
    int dfs(TreeNode* node, int sum) {
        if (node == nullptr)
            return 0;
        return (node->val == sum ? 1 : 0) + dfs(node->left, sum - node->val) + dfs(node->right, sum - node->val);
    }
};
// 22 ms

 

[LeetCode] Path Sum III

原文:http://www.cnblogs.com/immjc/p/7231524.html

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