112. Path Sum 113. Path Sum II

回溯法的典型

1、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.

/**
* 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:
bool hasPathSum(TreeNode* root, int sum) {
if(root==NULL)
return false;
if(root->val==sum&&root->left==NULL&&root->right==NULL)
return true;
return (hasPathSum(root->left,sum-root->val)||hasPathSum(root->right,sum-root->val));
}
};

2、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]
]

/**
* 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>>paths;
vector<int>path;
findPaths(root,sum,path,paths);
return paths;
}
void findPaths(TreeNode*root,int sum,vector<int>&path,vector<vector<int>>&paths)
{
if(!root)
return;
path.push_back(root->val);
if(!root->left&&!root->right&&root->val==sum)
paths.push_back(path);
findPaths(root->left,sum-root->val,path,paths);
findPaths(root->right,sum-root->val,path,paths);
path.pop_back();
}
};