LeetCode112 路径总和

前言

思路

比较简单的一个思路是遍历所有的路径，求和后再查找目标值。但是，最好的方法是一边遍历，一边比对。

代码

方法一：遍历后再查找

/**
 * 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:
    void findPath(TreeNode* node, vector<int>& path, vector<int>& res) {
        path.push_back(node -> val);
        if (node -> left == NULL && node -> right == NULL) {
            int sum = 0;
            for (int i = 0; i < path.size(); i++) {
                sum += path[i];
            }
            res.push_back(sum);
        }

        if (node -> left) {
            findPath(node -> left, path, res);
            path.pop_back();
        }

        if (node -> right) {
            findPath(node -> right, path, res);
            path.pop_back();
        }
    }
    bool hasPathSum(TreeNode* root, int targetSum) {
        vector<int> path;
        vector<int> result;
        if (root == NULL) return false;
        findPath(root, path, result);
        for (int i = 0; i < result.size(); i++) {
            if (result[i] == targetSum) {
                return true;
            }
        }
        return false;
    }
};

方法二：边遍历边查找

/**
 * 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:
    bool findPath(TreeNode* node, int count) {
        if (!node -> left && !node -> right && count == 0) return true;
        if (!node -> left && !node -> right) return false;

        if (node -> left) {
            count -= node -> left -> val;
            if (findPath(node -> left, count)) return true;
            count += node -> left -> val;
        }

        if (node -> right) {
            count -= node -> right -> val;
            if (findPath(node -> right, count)) return true;
            count += node -> right -> val;
        }

        return false;
    }
    bool hasPathSum(TreeNode* root, int targetSum) {
        if (root == NULL) return false;
        return findPath(root, targetSum - root -> val);
    }
};