Path and ii path-sum-ii
Title description
Given a binary tree and a value sum, please find the path where the sum of the node values from the root node to the leaf node is equal to sum,
for example:
Given the following binary tree, sum=22,
return
[↵ [5,4,11] ,2],↵ [5,8,4,5]↵]↵
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 andsum = 22,
return
[↵ [5,4,11,2],↵ [5,8,4,5]↵]↵
Problem-solving ideas
Backtracking
- Path: recorded in the track
Selection list: left and right subtrees that are not empty
End condition: reach the leaf node and meet the conditions - Note: Since you need to start from the subtree of the root node, you must first judge the root node separately, and then backtrack
/**
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
class Solution {
public:
vector<vector<int> > pathSum(TreeNode* root, int sum) {
vector<vector<int> > res;
vector<int> track;
/*由于要从根节点的子树开始判断，所以要先单独对根节点进行判断*/
if(root == NULL) return res; //根为空，直接返回
//先将根节点添加进路径中
track.push_back(root -> val);
backtrack(root, sum, track, res);
return res;
}
void backtrack(TreeNode* root, int& sum, vector<int>& track, vector<vector<int> >& res) {
if(root == NULL) return ;
if(root -> left == NULL && root -> right == NULL && sum - root -> val == 0) {
res.push_back(track);
return ;
}
//两个选择，只要子树不空就做选择
if(root -> left != NULL) {
sum -= root -> val;
//添加左子树
track.push_back(root -> left -> val);
backtrack(root -> left, sum, track, res);
sum += root -> val;
track.pop_back();
}
if(root -> right != NULL) {
sum -= root -> val;
//添加右子树
track.push_back(root -> right -> val);
backtrack(root -> right, sum, track, res);
sum += root -> val;
track.pop_back();
}
}
};
DFS
/**
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
class Solution {
public:
vector<vector<int> > pathSum(TreeNode* root, int sum) {
vector<vector<int> > res;
vector<int> track;
if(root == NULL) return res;
dfs(root, sum, track, res);
return res;
}
void dfs(TreeNode* root, int sum, vector<int> track, vector<vector<int> >& res) {
if(root == NULL) return ;
track.push_back(root -> val);
if(root -> left == NULL && root -> right == NULL && sum - root -> val == 0) {
res.push_back(track);
}
//进入下一层
if(root -> left != NULL) {
dfs(root -> left, sum - root -> val, track, res);
}
if(root -> right != NULL) {
dfs(root -> right, sum - root -> val, track, res);
}
}
};