Path Sum III

ID: 437; medium

Loading...LeetCode

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 to targetSum in the tree rooted at root. This include paths that starts at root and paths that do not start at root.

• findPath return the number of paths that have their sum equal to targetSum, but the paths must start at root.