LeetCode Oj 112. Path Sum solution report

LeetCode Oj 112. Path Sum solution report
112. Path SumMy SubmissionsQuestionTotal Accepted: 91133 Total Submissions: 295432 Difficulty: Easy

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 path5->4->11->2Which sum is 22.

 

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Obviously, deep search is simple and efficient. Note the leaf node definition. If left and right children are empty, it is the leaf node. If a node has a child, it cannot be used as a leaf.

My AC code

 

public class PathSum {static Boolean ok = false;/** * @param args */public static void main(String[] args) {TreeNode treeNode = new TreeNode(1);TreeNode t1 = new TreeNode(2);System.out.println(hasPathSum(treeNode, 1));System.out.println(hasPathSum(null, 0));treeNode.left = t1;System.out.println(hasPathSum(treeNode, 1));}public static boolean hasPathSum(TreeNode root, int sum) {if(root == null) return false;ok = false;        dfs(root, sum, 0);        return ok;    }public static void dfs(TreeNode root, int sum, int s) {if(ok == true) return;if(root.left == null && root.right == null) {if (s + root.val == sum) {ok = true;}return;}if(root.left != null) dfs(root.left, sum, s + root.val);if(root.right != null) dfs(root.right, sum, s + root.val);}}