Given a binary tree and a sum, find all root-to-leaf paths where each path‘s sum equals the given sum.
For example:sum
= 22,
5
/ 4 8
/ / 11 13 4
/ \ / 7 2 5 1
return
[ [5,4,11,2], [5,8,4,5] ]
判断二叉树中是否存在一条从根到叶子节点的路径，使得路径上的节点值之和等于所给的值，输出所有的路径
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
void dfs(TreeNode*root, vector<vector<int> >&result, vector<int>path, int pathSum, int sum){
path.push_back(root->val);
pathSum+=root->val;
if(root->left==NULL && root->right==NULL){
if(pathSum==sum) result.push_back(path);
return;
}
//搜索左子树
if(root->left)
dfs(root->left, result, path, pathSum, sum);
//搜索右子树
if(root->right)
dfs(root->right, result, path, pathSum, sum);
}
vector<vector<int> > pathSum(TreeNode *root, int sum) {
vector<vector<int> >result;
if(root==NULL)return result;
vector<int>path;
int pathSum=0;
dfs(root, result, path, pathSum, sum);
return result;
}
};LeetCode: Path Sum II [113],布布扣,bubuko.com
原文地址：http://blog.csdn.net/harryhuang1990/article/details/28587845
