题解 | #二叉树中和为某一值的路径(一)#

dfs+剪枝

/**
 * 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 = false;
    bool hasPathSum(TreeNode* root, int sum) {
        // write code here
        if(!root)return false; 
        dfs(root,sum,0);
        return flag;
    }
    void dfs(TreeNode* r,int sum,int ans){
        if(!r)return;
        if(flag)return;
        ans += r->val;
        if(ans == sum && r->left==NULL && r->right==NULL){
            flag = true;return;
        }
         dfs(r->left,sum,ans);
         dfs(r->right,sum,ans);
    }
    
};