112. Path Sum

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.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private boolean result = false;
    public boolean hasPathSum(TreeNode root, int sum) {
        if (root == null) return false;
        dfs(root, sum, 0);
        return result;
    }

    private void dfs(TreeNode root, int sum, int pathSum) {
        if (root == null) return;
        if (root.left == null && root.right == null) {
            if (pathSum + root.val == sum) {
                result = true;
                return;
            }
        }
        if (root.left != null) dfs(root.left, sum, pathSum + root.val);
        if (root.right != null) dfs(root.right, sum, pathSum + root.val);
    }
}

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