437. 路径总和 III

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题目

437. 路径总和 III 官方题解

方法1:自己写的

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int res;
    void dfs(TreeNode* root, bool flag, int sum, int targetSum) {
        if (!root) {
            return;
        }

        if (flag) {
            if (sum + root->val == targetSum) res ++;
            dfs(root->left, true, sum + root->val, targetSum);
            dfs(root->right, true, sum + root->val, targetSum);
        } else {
            if (root->val == targetSum) res ++;
            dfs(root->left, true, root->val, targetSum);
            dfs(root->left, false, 0, targetSum);
            dfs(root->right, true, root->val, targetSum);
            dfs(root->right, false, 0, targetSum);
        }
    }

    int pathSum(TreeNode* root, int targetSum) {
        dfs(root, false, 0, targetSum);
        return res;
    }
};

时间复杂度:O(n^2),n为节点数

空间复杂度:O(h),h为树的高度

127/130 test cases passed.

方法2:官方题解爆搜-比我的优雅

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class Solution {
public:
    int rootSum(TreeNode* root, long long targetSum) {
        if (!root) {
            return 0;
        }

        int ret = 0;
        if (root->val == targetSum) {
            ret++;
        }

        ret += rootSum(root->left, targetSum - root->val);
        ret += rootSum(root->right, targetSum - root->val);
        return ret;
    }

    int pathSum(TreeNode* root, int targetSum) {
        if (!root) {
            return 0;
        }

        int ret = rootSum(root, targetSum);
        ret += pathSum(root->left, targetSum);
        ret += pathSum(root->right, targetSum);
        return ret;
    }
};

时间复杂度:O(n^2),n为节点数

空间复杂度:O(h),h为树的高度

方法3:前缀和 + 哈希表

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class Solution {
public:
    unordered_map<long long, int> prefix;

    int dfs(TreeNode *root, long long curr, int targetSum) {
        if (!root) {
            return 0;
        }

        int ret = 0;
        curr += root->val;
        if (prefix.count(curr - targetSum)) {
            ret = prefix[curr - targetSum];
        }

        prefix[curr]++;
        ret += dfs(root->left, curr, targetSum);
        ret += dfs(root->right, curr, targetSum);
        prefix[curr]--;

        return ret;
    }

    int pathSum(TreeNode* root, int targetSum) {
        prefix[0] = 1;
        return dfs(root, 0, targetSum);
    }
};

时间复杂度:O(n),n为节点数

空间复杂度:O(h),h为树的高度