112题目如下：
Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.
For example:
Given the below binary tree and sum
= 22,
5
/
4 8
/ /
11 13 4
/
7 2 1
return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.
113题目如下：
Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.
For example:
Given the below binary tree and sum
= 22,
5
/
4 8
/ /
11 13 4
/ /
7 2 5 1
return
[ [5,4,11,2], [5,8,4,5] ]
112和113题目是类似的，都是找出等于给定值的路径，不过前者只看有没有，后者是要输出所有符合条件的路径。
112由于只要看有没有等于给定值的路径，所以可以用BFS，将每个树节点的val改为从根节点到当前节点的距离，这样改到树的叶子节点的时候，就可以判断是否有叶子节点的距离值符合要求。
113由于需要输出所有符合条件的路径，如果树节点的数据结构不可以改，那这就不适合用BFS，因为如果用BFS需要在每个树节点里维护一个到达当前节点经过的节点记录；所以这里改用DFS，通过修改一个存储路径的vector来统计所有满足条件的路径。
112题目代码
/**
* 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) {
queue<TreeNode*> q;
vector<TreeNode*> v;
if(root==NULL) return false;
q.push(root);
TreeNode* p=root;
while(q.size()!=0)
{
p=q.front();
q.pop();
if(p->left!=NULL)
{
p->left->val+=p->val;
q.push(p->left);
}
if(p->right!=NULL)
{
p->right->val+=p->val;
q.push(p->right);
}
if(p->left==NULL && p->right==NULL)
v.push_back(p);
}
for(int i=0;i<v.size();i++)
if(v[i]->val==sum) return true;
return false;
}
};
113题目代码
/**
* 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:
//用来递归的函数
void findSum(TreeNode* t,vector<int>& temp, vector<vector<int> >& record, int sum, int cur)
{
temp.push_back(t->val);
cur+=t->val;
//在叶节点找到了目标值
if(cur==sum && t->left==NULL && t->right==NULL) {
record.push_back(temp);
temp.erase(temp.end()-1);
cur-=t->val;
return ;
}
//查找左右节点
if(t->left!=NULL) findSum(t->left, temp, record, sum , cur);
if(t->right!=NULL) findSum(t->right, temp, record, sum , cur);
//返回前更新临时路径的记录和累加和
temp.erase(temp.end()-1);
cur-=temp[temp.size()-1];
return;
}
vector<vector<int>> pathSum(TreeNode* root, int sum) {
vector<vector<int> > record;//记录最终要返回的路径
if(root==NULL) return record;
vector<int> temp;//记录当前临时路径
findSum(root,temp,record,sum,0);
return record;
}
};