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.

For 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

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

Recursive solution

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

Non-recursive solution using Stack

public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if (root == null) return false; //Always check null first 
        Stack<TreeNode> trees = new Stack<>();
        Stack<Integer> values = new Stack<>();
        trees.push(root);
        values.push(sum);
        while (!trees.isEmpty()) {
            TreeNode node = trees.pop();
            int value = values.pop();
            //We cannot use `return node.val == sum` here because we are not in recursion, the program will stops immediately 
            if (node.left == null && node.right == null && node.val == value) return true;
            if (node.left != null) {
                trees.push(node.left);
                values.push(value - node.val);
            }
            if (node.right != null) {
                trees.push(node.right);
                values.push(value - node.val);
            }
        }
        return false;
    }
}

Analysis

This is a typical tree traversal problem
The recursive solution is concise and easy to implement
We check whether we reach to a leaf, if so, we compare the node.val with current sum
Otherwise, we recursively check the left and right of current node

The non-recursive solution looks complicated but it might be asked in the interview with a higher chance
We need to have two Stack because we need to store the current sum
The idea is similar to the recursive solution