[LeetCode] Path Sum (I & II & III)

112. Path Sum

Problem

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.

Solution Recursive

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

Update 2018-11

Solution Iterative

class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if (root == null) return false;
        Deque<TreeNode> nodestack = new ArrayDeque<>();
        Deque<Integer> sumstack = new ArrayDeque<>();
        nodestack.push(root);
        sumstack.push(sum);
        while (!nodestack.isEmpty()) {
            TreeNode curNode = nodestack.pop();
            int curSum = sumstack.pop();
            if (curNode.left == null && curNode.right == null && curNode.val == curSum) return true;
            if (curNode.right != null) {
                nodestack.push(curNode.right);
                sumstack.push(curSum-curNode.val);
            }
            if (curNode.left != null) {
                nodestack.push(curNode.left);
                sumstack.push(curSum-curNode.val);
            }
        }
        return false;
    }
}

Path Sum II

Problem

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum 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  5   1

Return:

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

Solution

class Solution {
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> res = new ArrayList<>();
        if (root == null) return res;
        dfs(root, sum, new ArrayList<Integer>(), res);
        return res;
    }
    private void dfs(TreeNode root, int sum, List<Integer> temp, List<List<Integer>> res) {
        if (sum == root.val && root.left == null && root.right == null) {
            temp.add(root.val);
            List<Integer> save = new ArrayList<>(temp); //important
            res.add(save);
            temp.remove(temp.size()-1);
        } else {
            temp.add(root.val);
            if (root.left != null) dfs(root.left, sum-root.val, temp, res);
            if (root.right != null) dfs(root.right, sum-root.val, temp, res);
            temp.remove(temp.size()-1);
        }
        
    }
}

Path Sum III

Problem

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

class Solution {
    public int pathSum(TreeNode root, int sum) {
        if (root == null) return 0;
        return helper(root, sum) + pathSum(root.left, sum) + pathSum(root.right, sum);
    }
    private int helper(TreeNode root, int sum) {        
        int count = 0;
        if (sum == root.val) count++;
        if (root.left != null) count += helper(root.left, sum-root.val);
        if (root.right != null) count += helper(root.right, sum-root.val);
        return count;
    }
}