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Path Sum II

时间:2015-03-22 15:00:00      阅读:156      评论:0      收藏:0      [点我收藏+]

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]
]

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    
    List<List<Integer>> re_list = new ArrayList<List<Integer>>();
    
    public int DFS(TreeNode root,int sum,List<Integer> list,int current) {
        if(root==null) return 0;
        List<Integer> tmp_list = new ArrayList<Integer>();
        tmp_list.addAll(list);
        if(root.left==null&&root.right==null&&(root.val+current)==sum) {
            tmp_list.add(root.val);
            re_list.add(tmp_list);
        }
        else{
            tmp_list.add(root.val);
            if(root.left!=null) DFS(root.left,sum,tmp_list,current+root.val);
            if(root.right!=null) DFS(root.right,sum,tmp_list,current+root.val);
        }
        return 1;
    }
    
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<Integer> list = new ArrayList<Integer>();
        DFS(root,sum,list,0);
        return re_list;
    }
}

 

Path Sum II

原文:http://www.cnblogs.com/mrpod2g/p/4357253.html

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