文章目录
一、题目
Given the root of a binary tree and an integer targetSum, return true if the tree has a root-to-leaf path such that adding up all the values along the path equals targetSum.
A leaf is a node with no children.
Example 1:
Input: root = 5,4,8,11,null,13,4,7,2,null,null,null,1, targetSum = 22
Output: true
Explanation: The root-to-leaf path with the target sum is shown.
Example 2:
Input: root = 1,2,3, targetSum = 5
Output: false
Explanation: There two root-to-leaf paths in the tree:
(1 --> 2): The sum is 3.
(1 --> 3): The sum is 4.
There is no root-to-leaf path with sum = 5.
Example 3:
Input: root = \[\], targetSum = 0
Output: false
Explanation: Since the tree is empty, there are no root-to-leaf paths.
Constraints:
The number of nodes in the tree is in the range 0, 5000.
-1000 <= Node.val <= 1000
-1000 <= targetSum <= 1000
二、题解
cpp
/**
* 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 traversal(TreeNode* root,int targetSum){
if(root->left == nullptr && root->right == nullptr){
return targetSum == 0;
}
if(root->left){
if(traversal(root->left,targetSum-root->left->val)) return true;
}
if(root->right){
if(traversal(root->right,targetSum-root->right->val)) return true;
}
return false;
}
bool hasPathSum(TreeNode* root, int targetSum) {
if(root == nullptr) return false;
return traversal(root,targetSum - root->val);
}
};