[LeetCode] Path Sum III

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。这条路径不需要从根节点开始，所以需要一个数组维护每个节点的值，使用前序遍历二叉树，然后利用递归将每一条从根节点出的的元素放入数组中，如果这条路径及其子路径的和等于sum，令cnt+1，然后将最后一个元素弹出。接着递归下一个节点。进行路径计算。直到递归结束返回cnt即可。

class Solution {
public:
    vector<int> preSum;
    int pathSum(TreeNode* root, int sum) {
        if (root == nullptr)
            return 0;
        preSum.push_back(root->val);
        int cnt = pathSum(root->left, sum) + pathSum(root->right, sum);
        int tmpSum = 0;
        for (int i = preSum.size() - 1; i >= 0; i--) {
            tmpSum += preSum[i];
            if (tmpSum == sum)
                cnt++;
        }
        preSum.pop_back();
        return cnt;
    }
};
// 13 ms

一个简洁的递归写法。

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