LeetCode 112. Path Sum

 112. 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.

 

Constraints:

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

	/**
	* 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 targetSum) {
	// edge
	if(!root)
	return false;
	// leaft
	if(!root->left and !root->right and root->val == targetSum)
	return true;
	// left
	if(root->left and hasPathSum(root->left, targetSum - root->val))
	return true;
	// right
	if(root->right and hasPathSum(root->right, targetSum - root->val))
	return true;
	return false;
	}
	};

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