LeetCode 113. Path Sum ii dfs Solution

LeetCode 113. Path Sum ii dfs Solution
113. Path Sum IIMy SubmissionsQuestionTotal Accepted: 72944 Total Submissions: 262389 Difficulty: Medium

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]]

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Show TagsShow Similar ProblemsHave you met this question in a real interview? YesNo

Given a target number, find all the root to the leaf of a tree and the path to this number. This question is relatively simple and can be solved by common DFS deep search. Use a list to store the number of paths. Note: After recursion, remove the last number added to the list.

My AC code

public class PathSumII {/** * @param args */public static void main(String[] args) {TreeNode n1 = new TreeNode(1);TreeNode n2 = new TreeNode(-2);TreeNode n3 = new TreeNode(-3);TreeNode n4 = new TreeNode(1);TreeNode n5 = new TreeNode(3);TreeNode n6 = new TreeNode(-2);TreeNode n7 = new TreeNode(-1);n1.left = n2;n1.right = n3;n2.left = n4;n2.right = n5;n3.left = n6;n4.left = n7;System.out.print(pathSum(n1, 2));}public static List
 
  > pathSum(TreeNode root, int sum) {List
  
   > result = new ArrayList
   
    >();if(root != null) dfs(root, sum, 0, new ArrayList
    
     (), result);return result;    }private static void dfs(TreeNode root, int sum, int s, ArrayList
     
       path, List
      
       > result) {if(root.left == null && root.right == null) {if(s + root.val == sum) {path.add(root.val);result.add((List
       
        ) path.clone());path.remove(path.size() - 1);}return;}path.add(root.val);if(root.left != null) dfs(root.left, sum, s + root.val, path, result);if(root.right != null) dfs(root.right, sum, s + root.val, path, result);path.remove(path.size() - 1);}}