Leetcode:Path Sum与Path Sum II

Path Sum：

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.
递归，实现代码：

/**
* Definition for binary tree
* 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 0;
if(root->left==NULL&&root->right==NULL)
return root->val==sum;
return hasPathSum(root->left,sum-root->val)||
hasPathSum(root->right,sum-root->val);
}
};

Path
Sum II：

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]
]

实现代码：

/**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<vector<int> > pathSum(TreeNode *root, int sum)
{
vector<vector<int> > res;
vector<int> path;
findPath(root, sum, path, res);
return res;
}

void findPath(TreeNode* root, int sum, vector<int> path, vector<vector<int> > & res)
{
if(root==NULL)
return;
path.push_back(root->val);
if(sum == root->val && root->left==NULL && root->right==NULL)
{
res.push_back(path);
path.pop_back();
return;
}
if (root->left)
findPath(root->left, sum-root->val,path,res);
if (root->right)
findPath(root->right,sum-root->val,path,res);
path.pop_back ();
}
};