Leetcode 112 Total path (2022.04.22)

Give you the root node of the binary tree root And an integer representing the sum of goals targetSum . Determine if there is Root node to leaf node The path of , The sum of the values of all nodes in this path is equal to the target and targetSum . If there is , return true ; otherwise , return false .

Leaf node A node without children .

Example 1：

Input ：root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
Output ：true
explain ： The root node to leaf node path equal to the target sum is shown in the figure above .
Example 2：

Input ：root = [1,2,3], targetSum = 5
Output ：false
explain ： There are two paths from root node to leaf node in the tree ：
(1 --> 2): And for 3
(1 --> 3): And for 4
non-existent sum = 5 The path from the root node to the leaf node .
Example 3：

Input ：root = [], targetSum = 0
Output ：false
explain ： Because the tree is empty , Therefore, there is no path from root node to leaf node .

Tips ：

The number of nodes in the tree is in the range [0, 5000] Inside
-1000 <= Node.val <= 1000
-1000 <= targetSum <= 1000

source ： Power button （LeetCode）
link ：https://leetcode-cn.com/problems/path-sum

Method 1 ： recursive

C++ Submission ：

/** * 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 hasPathSum(TreeNode *root, int sum) {
    
        if (root == nullptr) {
    
            return false;
        }
        if (root->left == nullptr && root->right == nullptr) {
    
            return sum == root->val;
        }
        return hasPathSum(root->left, sum - root->val) ||
               hasPathSum(root->right, sum - root->val);
    }
};