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  • Leetcode 112

    /**
     * Definition for a binary tree node.
     * struct TreeNode {
     *     int val;
     *     TreeNode *left;
     *     TreeNode *right;
     *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
     * };
     */
    class Solution {
    public:
        bool hasPathSum(TreeNode* root, int sum) {
            if(root == NULL) return false;
            int flag = 0;
            DFS(root,sum,flag);
            if(flag) return true;
            else return false;
        }
        void DFS(TreeNode* root, int sum,int& flag){
            if((root->left == NULL)&&(root->right == NULL)){
                if(root->val == sum)
                    flag = 1;
            }
            else if((root->left != NULL)&&(root->right == NULL)){
                root->left->val += root->val;
                DFS(root->left,sum,flag);
            }
            else if((root->left == NULL)&&(root->right != NULL)){
                root->right->val += root->val;
                DFS(root->right,sum,flag);
            }
            else if((root->left != NULL)&&(root->right != NULL)){
                root->left->val += root->val;
                DFS(root->left,sum,flag);
                root->right->val += root->val;
                DFS(root->right,sum,flag);
            }
            return;
        }
    };

    __

    简便写法:

    class Solution {
    public:
        bool hasPathSum(TreeNode* root, int sum) {
            if(!root) return false;
            if(root->left == NULL && root->right == NULL) return root->val == sum;
            return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val); //sum - root->val很灵性
        }
    };
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  • 原文地址:https://www.cnblogs.com/cunyusup/p/10348186.html
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