题目:

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]
]

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:
vector<vector<int>> pathSum(TreeNode* root, int sum) {
vector<vector<int> > ret;
// Terminal conditon 1 : to the null (not leaf node)
// even if sum equals zero at the time, becasue not leaf node
// so it is also not the available solution path
if (!root) return ret;
// Terminal condition 2 : leaf node
if ( !root->left && !root->right )
{
// if leaf node's val equal sum
if ( sum==root->val )
{
vector<int> tmp;
tmp.insert(tmp.begin(),root->val);
ret.push_back(tmp);
}
return ret;
}
// not leaf node : move forward to left and right
vector<vector<int> > l = Solution::pathSum(root->left, sum - root->val);
vector<vector<int> > r = Solution::pathSum(root->right, sum - root->val);
for ( size_t i = ; i<l.size(); ++i )
{
l[i].insert(l[i].begin(), root->val);
ret.push_back(l[i]);
}
for ( size_t i = ; i<r.size(); ++i )
{
r[i].insert(r[i].begin(), root->val);
ret.push_back(r[i]);
}
return ret;
}
};

tips:

与Path Sum思路类似(http://www.cnblogs.com/xbf9xbf/p/4508964.html

不同的地方是:只要不是leaf node,left和right两边的情况都要考虑。

============================================

并且有个地方需要缕清思路:如果递归时遇上root==NULL,直接返回空的ret是否合理?如果此时的sum==0呢?

root==NULL有以下三种情况

1. 如果整棵树是空树:即使sum==0也不满足条件

2. 某个非leaf node的left(或right)为空:则即使此时sum==0,往left(或right)方向走会得到root=NULL,因为此时root不是leaf node也不成立

============================================

第二次过这道题,DFS的思路比较清晰。头几次漏掉了onePath.push_back(root->val)这个语句,补上以后AC了。

/**
* 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> > pathSum(TreeNode* root, int sum)
{
vector<vector<int> > ret;
vector<int> onePath;
if ( root ) Solution::psum(root, sum, ret, onePath);
return ret;
}
static void psum(TreeNode* root, int sum, vector<vector<int> >& ret, vector<int> onePath)
{
if ( !root->left && !root->right )
{
if ( root->val==sum )
{
onePath.push_back(root->val);
ret.push_back(onePath);
return;
}
}
onePath.push_back(root->val);
if ( root->left ) Solution::psum(root->left, sum-root->val, ret, onePath);
if ( root->right ) Solution::psum(root->right, sum-root->val, ret, onePath);
}
};

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