递归

• 思路：
  • 终止条件是递归到根节点 root，剩余 target 与根节点值相等则路径存在，否则不存在；
  • 递归查找左子树或者右子树存在 target = target - root->val 的路径；

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