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.
bool hasPathSum(TreeNode* root, int sum) {
if(root == NULL)
return false;
if(root -> left == NULL && root -> right == NULL)
return root -> val == sum;
return hasPathSum(root -> left, sum - root -> val) || hasPathSum(root -> right, sum - root -> val);
}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.
vector<vector<int>> pathSum(TreeNode* root, int sum) {
vector<vector<int> > result;
vector<int> cur;
pathSum(root, sum, cur, result);
return result;
}
void pathSum(TreeNode* root, int gap, vector<int>& path, vector<vector<int> >& result)
{
if(root == NULL)
return;
path.push_back(root -> val);
if(root -> left == NULL && root -> right == NULL)
{
if(root -> val == gap)
result.push_back(path);
path.pop_back();
return;
}
pathSum(root -> left, gap - root -> val, path, result);
pathSum(root -> right, gap - root -> val, path, result);
path.pop_back();
}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.
法：利用了前序遍历，对于每个遍历到的节点进行处理，维护一个变量pre来记录之前路径之和，然后cur为pre加上当前节点值，如果cur等于sum，那么返回结果时要加1，然后对当前节点的左右子节点调用递归函数求解，最终结果为从根出发及从其左右子树分别出发路径数目的总和。
int pathSum(TreeNode* root, int sum) {
if (!root)
return 0;
return sumUp(root, 0, sum) + pathSum(root->left, sum) + pathSum(root->right, sum);
}
int sumUp(TreeNode* node, int pre, int& sum) {
if (!node)
return 0;
int cur = pre + node -> val;
return (cur == sum) + sumUp(node->left, cur, sum) + sumUp(node->right, cur, sum);
}62. Unique Paths
A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).
The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).
How many possible unique paths are there?
Above is a 7 x 3 grid. How many possible unique paths are there?
Note: m and n will be at most 100.
法： dp[i][j] = dp[i- 1][j] + dp[i][j - 1];
int uniquePaths(int m, int n) {
if(m == 0 || n == 0)
return 1;
vector<vector<int>> dp(m, vector<int>(n));
for(int i = 0; i < m; i ++)
dp[i][0] = 1;
for(int j = 0; j < n; j ++)
dp[0][j] = 1;
for(int i = 1; i < m; i ++)
for(int j = 1; j < n; j ++)
dp[i][j] = dp[i- 1][j] + dp[i][j - 1];
return dp[m-1][n-1];
}63. Unique Paths II
A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).
The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).
Now consider if some obstacles are added to the grids. How many unique paths would there be?
An obstacle and empty space is marked as 1 and 0 respectively in the grid.
Note: m and n will be at most 100.
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
int m = obstacleGrid.size();
int n = obstacleGrid[0].size();
vector<vector<int> > dp(m, vector<int>(n,0));
for(int i = 0; i < m; i ++)
if(obstacleGrid[i][0] == 1)
break;
else
dp[i][0] = 1;
for(int j = 0; j < n; j ++)
if(obstacleGrid[0][j] == 1)
break;
else
dp[0][j] = 1;
for(int i = 1; i < m; i ++)
for(int j = 1; j < n; j++)
if(obstacleGrid[i][j] != 1)
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
return dp[m - 1][n - 1];
}