LeetCode题解：Path Sum I and 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.

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]
]

思路：

递归post-order访问树的同时求和即可。这里采用了另外一个办法，与上述相似。

题解：

/**
* 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 == nullptr)
return false;

if (root->left == nullptr && root->right == nullptr)
{
if (sum == root->val)
return true;
else
return false;
}
else
return (root->left == nullptr ?
false : hasPathSum(root->left, sum - root->val)) ||
(root->right == nullptr ?
false : hasPathSum(root->right, sum - root->val));

}
};
思路：
preorder访问树，迭代进行。用一个vector保存路径。

/**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<vector<int>> scratch;
vector<int> history;

void traverse(TreeNode* node, int sum)
{
history.push_back(node->val);
if (node->left == nullptr && node->right == nullptr &&
accumulate(begin(history), end(history), 0) == sum)
scratch.push_back(history);
if (node->left != nullptr)
traverse(node->left, sum);
if (node->right != nullptr)
traverse(node->right, sum);
history.pop_back();
}

vector<vector<int> > pathSum(TreeNode *root, int sum) {
scratch.clear();
if (root == nullptr)
return scratch;
traverse(root, sum);
return scratch;
}
};