Paste_Image.png

My code:

import java.util.ArrayList;
import java.util.List;

/**
 * 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) {
        ArrayList<List<Integer>> result = new ArrayList<List<Integer>>();
        if (root == null)
            return result;
        hasPathSum(root, 0, sum, new ArrayList<Integer>(), result);
        return result;
    }
    
    private void hasPathSum(TreeNode root, int sum, int totalSum, ArrayList<Integer> path,
                                ArrayList<List<Integer>> result) {
        if (root == null) {
            return;
        }
        path.add(root.val);
        if (root.left == null && root.right == null) {
            if (sum + root.val == totalSum)
                result.add(new ArrayList<Integer>(path));
            path.remove(path.size() - 1);   
            return;
        }
        else {
            hasPathSum(root.left, sum + root.val, totalSum, path, result);
            hasPathSum(root.right, sum + root.val, totalSum, path, result);
            path.remove(path.size() - 1);
        }
    }
}

My test result:

这道题目没有什么难度。
感觉还是用到了以前permutation系列题的思想，把已经用完的了结点删除。便于下一次深度遍历。DFS

**
总结， Tree DFS
**

Anyway, Good luck, Richardo!

My code:

/**
 * 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>> ret = new ArrayList<List<Integer>>();
        if (root == null) {
            return ret;
        }
        List<Integer> group = new ArrayList<Integer>();
        helper(root, 0, sum, ret, group);
        return ret;
    }
    
    private void helper(TreeNode root, int curr, int sum, List<List<Integer>> ret, List<Integer> group) {
        if (root.left == null && root.right == null) {
            if (root.val + curr == sum) {
                group.add(root.val);
                ret.add(new ArrayList<Integer>(group));
                group.remove(group.size() - 1);
            }
            else {
                return;
            }
        }
        else {
            group.add(root.val);
            if (root.left != null) {
                helper(root.left, root.val + curr, sum, ret, group);
            }
            
            if (root.right != null) {
                helper(root.right, root.val + curr, sum, ret, group);
            }
            group.remove(group.size() - 1);
        }
    }
}

注意力没集中，没想到这道题目做了这么久。真他妈了隔壁的。。

其实就是 permutation的思想。只是形式稍微变一下。

Anyway, Good luck, Richardo! -- 08/28/2016