leetcode -- 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]
]
 1 /**
 2  * Definition for binary tree
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public ArrayList<ArrayList<Integer>> pathSum(TreeNode root, int sum) {
12         // Start typing your Java solution below
13         // DO NOT write main() function
14         ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
15         if(root == null){
16             return result;            
17         }
18         ArrayList<Integer> seqs = new ArrayList<Integer>();
19         int curSum = 0;
20         
21         generatePathSum(root, result, seqs, curSum, sum);
22         return result;
23     }
24     
25     public void generatePathSum(TreeNode root, ArrayList<ArrayList<Integer>> result,
26         ArrayList<Integer> seqs, int curSum, int expectedSum){
27             curSum += root.val;
28             seqs.add(root.val);
29             
30             boolean isLeaf = root.left == null && root.right == null;
31             if(curSum == expectedSum && isLeaf){
32                 ArrayList<Integer> tmp = new ArrayList<Integer>();
33                 tmp.addAll(seqs);
34                 result.add(tmp);
35             }
36             
37             if(root.left != null){
38                 generatePathSum(root.left, result, seqs, curSum, expectedSum);
39             }
40             
41             if(root.right != null){
42                 generatePathSum(root.right, result, seqs, curSum, expectedSum);
43             }
44             
45             seqs.remove(seqs.size() - 1);
46             
47         }
48 }

 refactor code:

 1 public ArrayList<ArrayList<Integer>> pathSum(TreeNode root, int sum) {
 2         // Start typing your Java solution below
 3         // DO NOT write main() function
 4         ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
 5         if(root == null){
 6             return result;            
 7         }
 8         ArrayList<Integer> seqs = new ArrayList<Integer>();
 9         int curSum = 0;
10         
11         generatePathSum(root, result, seqs, curSum, sum);
12         return result;
13     }
14     
15     public void generatePathSum(TreeNode root, ArrayList<ArrayList<Integer>> result,
16         ArrayList<Integer> seqs, int curSum, int expectedSum){
17             curSum += root.val;
18             seqs.add(root.val);
19             
20             boolean isLeaf = root.left == null && root.right == null;
21             if(curSum == expectedSum && isLeaf){
22                 ArrayList<Integer> tmp = new ArrayList<Integer>();
23                 tmp.addAll(seqs);
24                 result.add(tmp);
25                 seqs.remove(seqs.size() - 1);
26                 return;
27             }
28             
29             if(root.left != null){
30                 generatePathSum(root.left, result, seqs, curSum, expectedSum);
31             }
32             
33             if(root.right != null){
34                 generatePathSum(root.right, result, seqs, curSum, expectedSum);
35             }
36             
37             seqs.remove(seqs.size() - 1);
38             
39         }

 

posted @ 2013-08-10 23:35  feiling  阅读(205)  评论(0)  编辑  收藏  举报