题解 | #二叉树中和为某一值的路径(一)#

/**

• struct TreeNode {
• int val;
• struct TreeNode *left;
• struct TreeNode *right;
• }; */

class Solution { public: /** * * @param root TreeNode类 * @param sum int整型 * @return bool布尔型 / bool flag = 0; bool hasPathSum(TreeNode root, int sum) { // write code here if(!root) return false; dfs(root, sum); return flag; } void dfs(TreeNode* root, int sum) {

    if(root -> val == sum && root -> left == nullptr && root -> right == nullptr)
    {
        flag = true;
        return;
    }
    if(root -> left) dfs(root -> left, sum - (root -> val));
    if(root -> right) dfs(root -> right, sum - (root -> val));
    return;
}

};