LeetCode No.437 Path Sum III

You are given a binary tree in which each node contains an integer value.

Find the number of paths that sum to a given value.

The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).

The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.

Example:

root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8

10
/  \
5   -3
/ \    \
3   2   11
/ \   \
3  -2   1

Return 3. The paths that sum to 8 are:

1.  5 -> 3
2.  5 -> 2 -> 1
3. -3 -> 11

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这道题目的大意是：找出二叉树中和为sum的所有路径（路径起点任意）

本节点路径值 = 本身 + （上一节点的路径值+本身），再利用深搜（DFS）或者宽搜（BFS）就行。

附上代码：

/**
* 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:
int pathSum(TreeNode* root, int sum) {
if ( root == NULL )
return 0 ;
int ans = 0 ;
vector < pair <TreeNode*,vector<int> > > v ;//节点，路径所有可能值
vector <int> num ( 1 , root -> val ) ;
v.push_back ( make_pair ( root , num ) ) ;
while ( ! v.empty() )
{
TreeNode* r = v.back().first ;
num = v.back().second ;
v.pop_back () ;
for ( int i = 0 ; i < num.size() ; i ++ )//找出所有符合条件的路径
if ( num[i] == sum )
ans ++ ;
if ( r -> left )
{
int numb = r -> left -> val ;
vector <int> temp ( 1 , numb ) ;//路径值+=本身
for ( int i = 0 ; i < num.size() ; i ++ )//路径值+=上一节点的路径值+本身
temp.push_back ( num[i] + numb ) ;
v.push_back ( make_pair ( r -> left , temp ) ) ;
}
if ( r -> right )
{
int numb = r -> right -> val ;
vector <int> temp ( 1 , numb ) ;//路径值+=本身
for ( int i = 0 ; i < num.size() ; i ++ )//路径值+=上一节点的路径值+本身
temp.push_back ( num[i] + numb ) ;
v.push_back ( make_pair ( r -> right , temp ) ) ;
}
}
return ans ;
}
};