• 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]
    ]

    /**
     * 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;
        }
    }
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  • 原文地址:https://www.cnblogs.com/mrpod2g/p/4357253.html
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