112,113,114. Path Sum I && II & III

Path Sum I

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

Note: A leaf is a node with no children.

Example:

Given the below binary tree and sum = 22,

      5
     / \
    4   8
   /   / \
  11  13  4
 /  \      \
7    2      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if (root == null) return false;
        if (root.left == null && root.right == null && root.val == sum) return true;
        return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
    }
}

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,

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \    / \
        7    2  5   1

return

[
   [5,4,11,2],
   [5,8,4,5]
]

/**
 * 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>> all = new ArrayList<>();
        helper(root, new ArrayList<Integer>(), all, 0, sum);
        return all;
    }
    
    private void helper(TreeNode root, List<Integer> list, List<List<Integer>> all, int currSum, int sum) {
        // exit condition
        if (root == null) return;
        
        list.add(root.val);
        currSum += root.val;
        
        if (currSum == sum && root.left == null && root.right == null) {
            all.add(new ArrayList<Integer>(list));
        }
        
        helper(root.left, list, all, currSum, sum);
        helper(root.right, list, all, currSum, sum);
        list.remove(list.size() - 1);
    }
}

Path Sum III

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

public class Solution {
    public int pathSum(TreeNode root, int sum) {
        int[] arr = new int[1];
        preOrder(root, sum, arr);
        return arr[0];
    }

    public void preOrder(TreeNode root, int sum, int[] count) {
        if (root == null) return;
        printSums(root, sum, 0, count);
        preOrder(root.left, sum, count);
        preOrder(root.right, sum, count);
    }

    public void printSums(TreeNode root, int sum, int currentSum, int[] count) {
        if (root == null) return;
        currentSum += root.val;
        if (currentSum == sum) {
            count[0]++;
        }
        printSums(root.left, sum, currentSum, count);
        printSums(root.right, sum, currentSum, count);
    }
}