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    Binary Tree Path Sum II

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    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]
    ]
    

    Approach 1:

    /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    public class Solution {
        private List<List<Integer>> resultList = new ArrayList<List<Integer>>();
        
        public void pathSumInner(TreeNode root, int sum, Stack<Integer>path) {
            path.push(root.val);
            if(root.left == null && root.right == null)
                if(sum == root.val) resultList.add(new ArrayList<Integer>(path));
            if(root.left!=null) pathSumInner(root.left, sum-root.val, path);
            if(root.right!=null)pathSumInner(root.right, sum-root.val, path);
            path.pop();
        }
        
        public List<List<Integer>> pathSum(TreeNode root, int sum) {
            if(root==null) return resultList;
            Stack<Integer> path = new Stack<Integer>();
            pathSumInner(root, sum, path);
            return resultList;
        }
    }

    Approach 2:

    /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    public class Solution {
      public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> result = new ArrayList<>();
        helper(root, sum, result, new ArrayList<>());
    
        return result;
      }
    
      private void helper(TreeNode root, int sum, List<List<Integer>> result, List<Integer> list) {
        if (root == null) {
          return;
        }
        list.add(root.val);
        if (root.left == null && root.right == null && root.val == sum) {
          result.add(new ArrayList<>(list));
            list.remove(list.size()-1);
          return;
        }
        helper(root.left, sum - root.val, result, list);
        helper(root.right, sum - root.val, result, list);
        list.remove(list.size()-1);
      }
    }

     

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