437.Path Sum III

Tags: Tree Easy

Links: https://leetcode.com/problems/path-sum-iii/

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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

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/**
 * 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 {
public:
    int pathSum(TreeNode* root, int sum) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        if (!root) return 0;
        queue<TreeNode *> q;
        q.push(root);
        int res = 0;

        while (!q.empty()) {
            int n = q.size();
            for (int i = 0; i < n; ++i) {
                TreeNode *tmp = q.front(); q.pop();

                int cnt = 0;
                helper(tmp, sum, cnt);
                res += cnt;

                if (tmp -> left) q.push(tmp -> left);
                if (tmp -> right) q.push(tmp -> right);
            }
        }

        return res;
    }

    void helper(TreeNode *root, int sum, int &cnt)
    {
        if (root -> val == sum) {
            cnt += 1;
        }

        if (root -> left) helper(root -> left, sum - root -> val, cnt);
        if (root -> right) helper(root -> right, sum - root -> val, cnt);
    }
};

用helper代表以当前根节点为起始点的式路径和为sum的种类数，用一个cnt的引用来记录。考虑到每一个节点都可能被当作根节点，所以相当于二叉树的层序遍历。