Path Sum III
ID: 437; medium
Solution 1 (Java)
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public int pathSum(TreeNode root, int targetSum) {
if (root == null)
return 0;
return pathSum(root.left, targetSum)
+ pathSum(root.right, targetSum) + findPath(root, targetSum);
}
private int findPath(TreeNode root, int targetSum) {
if (root == null)
return 0;
int res = 0;
if (root.val == targetSum)
res += 1;
res += findPath(root.left, targetSum - root.val);
res += findPath(root.right, targetSum - root.val);
return res;
}
}
Notes
pathSum
returns the number of paths that have their sum equal totargetSum
in the tree rooted atroot
. This include paths that starts atroot
and paths that do not start atroot
.findPath
return the number of paths that have their sum equal totargetSum
, but the paths must start atroot
.
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