【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）
112. 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.
Note: A leaf is a node with no children.
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.
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
/*
只要求返回true或false,因此不需要记录路径
*/
class Solution
{
public:
bool hasPathSum(TreeNode* root, int sum)
{
if(root == nullptr) return false;
if(root->left == nullptr && root->right == nullptr) //叶子结点
return sum == root->val;
int newsum = sum - root->val;
return hasPathSum(root->left, newsum) || hasPathSum(root->right, newsum);
}
};
113. 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.
Note: A leaf is a node with no children.
Example:
Given the below binary tree and sum = 22,
Return:
[
[5,4,11,2],
[5,8,4,5]
]
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
/*
求所有和等于某数的路径
*/
#include <numeric> //算容器类元素和可用accumulate函数
class Solution
{
public:
vector<vector<int>> pathSum(TreeNode* root, int sum)
{
vector<vector<int>> result;
vector<int> path;
path_sum(root, path, result, sum);
return result;
}
private:
void path_sum(TreeNode* root, vector<int>& path, vector<vector<int>>& result, int gap)
{
if(root == nullptr)
return; //递归出口
else
path.push_back(root->val); //存储结点元素到path
if(root->left == nullptr && root->right == nullptr) //叶子结点时push path到结果向量中
{
if(gap == root->val) result.push_back(path); //如果该path和为sum则push到结果向量中（这里用sum累减路径上的元素，得到gap与路径上最后一个元素比较，节省时间，如果得到path再accumulate，则会造成不同路径间的重复计算）
// return; //递归出口，到叶结点后退出，（不能写这句，还需运行到结尾进行pop）
}
path_sum(root->left, path, result, gap - root->val); //沿深度方向遍历
path_sum(root->right, path, result, gap - root->val);
path.pop_back();//删除最后一个元素，腾出空间(本函数中只push了一次，故只需pop一次)
}
};