113. Path Sum II

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]
]

Analysis:

Pre-order traverse the tree, use a global variable to store path list, remember to pop one when recurse.

Complexity:

Time: O(n)

Space: O(logn)

Code:

/**
 * 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<Integer> path = new ArrayList<Integer>();
    List<List<Integer>> returnList;
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        returnList = new ArrayList<List<Integer>>();
        if(root == null)
        {
            return returnList;
        }

        findPath(root, sum);
        return returnList;
    }

    private void findPath(TreeNode root, int sum)
    {
        if(root == null)
        {
            return;
        }
        path.add(root.val);
        if(root.left == null && root.right == null && sum - root.val == 0)
        {
            returnList.add(new ArrayList<Integer>(path));
        }
        else
        {
            findPath(root.left, sum - root.val);
            findPath(root.right, sum - root.val);
        }
        path.remove(path.size()-1);

    }
}