Path Sum
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
C++ Solution
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bool hasPathSum(TreeNode* root, int targetSum) {
if (root == nullptr) {
return false;
}
int newSum = targetSum - root->val;
if (isLeaf(root) && newSum ==0 ) {
return true;
}
return hasPathSum(root->left, newSum) ||
hasPathSum(root->right, newSum);
}
bool isLeaf(TreeNode *node) {
return node != nullptr &&
node->left == nullptr && node->right == nullptr;
}
