力扣-二叉树-112 路径总和

思路

使用先序遍历，回溯不同节点的值

代码

class Solution {
public:

    bool havePath = false;

    void prefix(TreeNode* node, int target, int result){
        if(node->left == nullptr && node->right == nullptr && result + node->val == target){
            havePath = true;
            return ;
        }
        if(havePath) return;
        if(node->left){
            result += node->val;
            prefix(node->left, target, result);
            result -= node->val;
        }
        if(node->right){
            result += node->val;
            prefix(node->right, target, result);
            result -= node->val;
        }
    }

    bool hasPathSum(TreeNode* root, int targetSum) {
        if(root == nullptr) return false;
        prefix(root, targetSum, 0);
        return havePath;
    }
};