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]
]


方法1：只要在每个树节点中加入一个父节点，就可以用上一题的解法了。
方法2：dfs

方法1的代码：
 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  
12     class TreeNodePlus {
13         TreeNode node;
14         TreeNodePlus father;
15         TreeNodePlus(TreeNode x,TreeNodePlus f) {
16             node=x;
17             father=f;
18         }
19     }
20 
21     List<List<Integer>> paths = new ArrayList<List<Integer>>();
22 
23     public List<List<Integer>> pathSum(TreeNode root, int sum) {
24         TreeNodePlus newRoot = new TreeNodePlus(root, null);
25         hasPathSum(newRoot, sum);
26         return paths;
27     }
28 
29     public void hasPathSum(TreeNodePlus root, int sum){
30         if (root.node == null) {
31             return ;
32         }
33         if (root.node.val==sum && root.node.left==null && root.node.right==null) {
34             ArrayList<Integer> list = new ArrayList<Integer>();
35             TreeNodePlus father = root;
36             while (father != null) {
37                 list.add(0,father.node.val);
38                 father = father.father;
39 
40             }
41             paths.add(list);
42             return;
43         }
44 
45         hasPathSum(new TreeNodePlus(root.node.left,root), sum-root.node.val) ;
46         hasPathSum(new TreeNodePlus(root.node.right,root), sum-root.node.val);
47     }
48 
49 }

 

　　方法2的代码

 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  
12 
13 
14     List<List<Integer>> paths = new ArrayList<List<Integer>>();
15     public List<List<Integer>> pathSum(TreeNode root, int sum) {
16 
17         List<Integer> list = new ArrayList<Integer>();
18         dfs(root, sum, 0, list);
19         return paths;
20     }
21     void dfs(TreeNode root, int sum, int curr, List<Integer> list) {
22         if (root == null) {
23             return;
24         }
25 
26         if (root.left == null && root.right == null) {
27             if (curr + root.val == sum) {
28                 list.add(root.val);
29                 paths.add(list);
30 
31             }
32             return;
33         }
34 
35         list.add(root.val);
36         dfs(root.left, sum, curr + root.val, new ArrayList<Integer>(list));
37 
38         dfs(root.right, sum, curr + root.val, new ArrayList<Integer>(list));
39     }
40 
41 }

 

posted @ 2014-12-15 23:40  birdhack  阅读(132)  评论(0)  编辑  收藏  举报