Question
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.
Example:
root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8
10
/ \
5 -3
/ \ \
3 2 11
/ \ \
3 -2 1
Return 3. The paths that sum to 8 are:
1. 5 -> 3
2. 5 -> 2 -> 1
3. -3 -> 11
Solution
TODO
Code
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int pathSum(TreeNode root, int sum) {
// path sum start from all nodes below root
if(root == null) return 0;
return pathSum(root.left, sum) +
pathSum(root.right, sum) +
pathSumFrom(root, sum);
}
public int pathSumFrom(TreeNode root, int sum) {
// path sum start from the specific node
if(root == null) return 0;
int count = 0;
if(root.val == sum) count++;
count += pathSumFrom(root.left, sum - root.val);
count += pathSumFrom(root.right, sum - root.val);
return count;
}
}
class Solution {
int count = 0;
public int pathSum(TreeNode root, int sum) {
dfs(root, sum);
return count;
}
private List<Integer> dfs(TreeNode root, int sum) {
List<Integer> results = new ArrayList<>();
if(root == null) return results;
int val = root.val;
List<Integer> left = dfs(root.left, sum);
List<Integer> right = dfs(root.right, sum);
results.add(val);
for(int v: left) results.add(v+val);
for(int v: right) results.add(v+val);
for(int v: results) if(v == sum) count++;
return results;
}
}
Performance
O(number of nodes)