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]
]

一刷
題解：用DFS

public List<List<Integer>> pathSum(TreeNode root, int sum){
    List<List<Integer>> result  = new LinkedList<List<Integer>>();
    List<Integer> currentResult  = new LinkedList<Integer>();
    pathSum(root,sum,currentResult,result);
    return result;
}

public void pathSum(TreeNode root, int sum, List<Integer> currentResult,
        List<List<Integer>> result) {

    if (root == null)
        return;
    currentResult.add(new Integer(root.val));
    if (root.left == null && root.right == null && sum == root.val) {
        result.add(new LinkedList(currentResult));
        currentResult.remove(currentResult.size() - 1);//don't forget to remove the last integer
        return;
    } else {
        pathSum(root.left, sum - root.val, currentResult, result);
        pathSum(root.right, sum - root.val, currentResult, result);
    }
    currentResult.remove(currentResult.size() - 1);
}

二刷
思路同上。注意，最后的結束點需要是葉子結點泳姐。

/**
 * 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>> res = new ArrayList<>();
        List<Integer> list = new ArrayList<>();
        if(root == null) return res;
        pathSum(res, list, root, sum);
        return res;
        
    }
    
    private void pathSum(List<List<Integer>> res, List<Integer> list, TreeNode root, int sum){
        if(root == null) return;
        list.add(root.val);
        if(root.val == sum && root.left == null && root.right == null){
            res.add(new ArrayList<Integer>(list));
            list.remove(list.size()-1);
            return;
        }
        else{
            pathSum(res, list, root.left, sum-root.val);
            pathSum(res, list,root.right, sum-root.val);
            list.remove(list.size()-1);//remove the current root
        }
        
        
    }
}

三刷
DFS

/**
 * 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>> res = new ArrayList<>();
        if(root == null) return res;
        dfs(res, new ArrayList<>(), root, sum);
        return res;
    }
    
    private void dfs(List<List<Integer>> res, List<Integer> list, TreeNode root, int sum){
        if(root == null) return;
        list.add(root.val);
        if(root.val == sum && root.left == null && root.right == null){
            res.add(new ArrayList<>(list));
            list.remove(list.size()-1);
            return;
        }
        dfs(res, list, root.left, sum-root.val);
        dfs(res, list, root.right, sum-root.val);
        list.remove(list.size()-1);
    }
}