算法-01-二叉树根节点到叶子节点和为指定值的路径

描述

给定一个二叉树和一个值sum，请找出所有的根节点到叶子节点的节点值之和等于sum 的路径， 例如： 给出如下的二叉树，sum=22

Java 

import java.util.*;

/*
 * public class TreeNode {
 *   int val = 0;
 *   TreeNode left = null;
 *   TreeNode right = null;
 * }
 */

public class Solution {
    /**
     * 
     * @param root TreeNode类 
     * @param sum int整型 
     * @return int整型ArrayList<ArrayList<>>
     */
    ArrayList<ArrayList<Integer>> results= new ArrayList<>();
    
    public ArrayList<ArrayList<Integer>> pathSum (TreeNode root, int sum) {
        // write code here
        ArrayList<Integer> path =new ArrayList<>();
        if(root==null)
            return results;
        dfs(root,0,sum,path);
        return results;
    }
    
    public void dfs(TreeNode root,int nowSum, int sum, ArrayList<Integer> list){
        list.add(root.val);
        if(root.right==null&&root.left==null&&sum==nowSum+root.val){
            results.add(new ArrayList<>(list));
            return;
        }
        if(root.left!=null){
            dfs(root.left,nowSum+root.val,sum,list);
            list.remove(list.size()-1);
        }
        if(root.right!=null){
            dfs(root.right, nowSum+root.val,sum,list);
            list.remove(list.size()-1);
        }
    }
}

 Python 

# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

#
# 
# @param root TreeNode类 
# @param sum int整型 
# @return int整型二维数组
#
class Solution:
    def pathSum(self , root , sum ):
        # write code here
        path = []
        paths = []
        self.dfs(root,sum,path,paths)
        return paths
        
    def dfs(self, root: TreeNode, sum :int, path: list, paths: list):
        if root is None:
            return None
        path.append(root.val)
        if root.left is None and root.right is None and sum==root.val:
            paths.append(path.copy())
        else:
            self.dfs(root.left,sum-root.val,path,paths)
            self.dfs(root.right,sum-root.val,path,paths)
        path.pop(len(path)-1)