路径总和I
/**
* 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==nullptr) return false;
sum-=root->val;
if(root->left==nullptr&&root->right==nullptr&&sum==0) return true;
return hasPathSum(root->left,sum)||hasPathSum(root->right,sum);
}
};
路径总和II
/**
* 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:
vector<vector<int>> ans;//最终答案,默认未修改为全局变量,为空
vector<int> path;//路径
vector<vector<int>> pathSum(TreeNode* root, int sum) {
dfs(root,sum);//开始深度优先搜索
return ans;//返回答案
}
void dfs(TreeNode * root,int sum){
if(root==nullptr) return;//若为空节点,则直接结束
path.push_back(root->val);
sum-=root->val;//修改当前的sum值,要首先进行修改sum的值,否则将会没有合法答案出现。
if(root->left==nullptr&&root->right==nullptr&&sum==0){
ans.push_back(path);//判定为最终的叶子节点,则满足条件者进行压入容器末尾
}
dfs(root->left,sum);//向左边子树进行遍历
dfs(root->right,sum);//向右边子树进行遍历
path.pop_back();//不满足者直接弹出
}
};
路径总和III
(continuing)