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.
For example: Given the below binary tree and sum = 22,
return
:::tip Idea Recursion :::
Solution:¶
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<List<Integer>> pathSum(TreeNode root, int sum) {
List<List<Integer>> result = new LinkedList<List<Integer>>();
process(root, sum, new LinkedList<Integer>(), result);
return result;
}
private void process(TreeNode node, int sum, List<Integer> curr, List<List<Integer>> result){
if (node == null) {
return;
}
curr.add(node.val);
if (node.left == null && node.right == null & node.val == sum) {
result.add(new LinkedList<Integer>(curr));
}
process(node.left, sum - node.val, curr, result);
process(node.right, sum - node.val, curr, result);
curr.remove(curr.size() - 1);
}
}
// Runtime: 9 ms
class Solution {
public:
vector<vector<int>> pathSum(TreeNode* root, int sum) {
vector<vector<int>> ans;
vector<int> curr;
pathSum(root, sum, curr, ans);
return ans;
}
private:
void pathSum(TreeNode* root, int sum, vector<int>& curr, vector<vector<int>>& ans) {
if(root==nullptr) return;
if(root->left==nullptr && root->right==nullptr) {
if(root->val == sum) {
ans.push_back(curr);
ans.back().push_back(root->val);
}
return;
}
curr.push_back(root->val);
int new_sum = sum - root->val;
pathSum(root->left, new_sum, curr, ans);
pathSum(root->right, new_sum, curr, ans);
curr.pop_back();
}
};