[PYTHON] Reviewing the wooden structure and challenging DFS

Good evening (* ´ω ｀)

How are you doing at the end of the year? Because I was tipsy and somehow challenged DFS I will make a note of it. ('◇') ゞ

DFS articles have a lot of experts I will omit the explanation w

I wrote it somehow and tried to move it Share the results.

Tree_DFS.py

from collections import deque

class Node:
    def __init__(self,val,left = None,right = None,):
        self.val = val
        self.left = left
        self.right = right

class BinTree:
    def __init__(self):
        self.root = None
        
    def add(self,val):                    
        def add_node(node,val):
            runner = node
            if val < runner.val:
                if not runner.left:
                    runner.left = Node(val)
                else:
                    add_node(runner.left,val)
            if val > runner.val:
                if not runner.right:
                    runner.right = Node(val)
                else:
                    add_node(runner.right,val)
                    
        if not self.root:
            self.root = Node(val)
        else:
            add_node(self.root,val) 

    def show(self):
        
        if not self.root:
            print("None")
            return
        
        def DFS(node):
            print(node.val)
            
            if node.left:
                DFS(node.left)
            if node.right:
                DFS(node.right)
                
        return DFS(self.root)

  #case 1#
### TreeImage ###
#       5       #
#      / \      #
#     4   10    #
#    /    /\    #
#   3    6  11  #
#################


T = BinTree()
T.add(5)
T.add(4)
T.add(10)
T.add(6)
T.add(11)
T.add(3)
T.show()

Execution result.py

  #case 1#
### TreeImage ###
#       5       #
#      / \      #
#     4   10    #
#    /    /\    #
#   3    6  11  #
#################

5
4
3
10
6
11

It seems to be working implicitly. Is it good, is it good? ..

Let's put it into practice. 103. Binary Tree Zigzag Level Order Traversal Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).

For example: Given binary tree [3,9,20,null,null,15,7],

Tree_image.py

    3
   / \
  9  20
    /  \
   15   7

return its zigzag level order traversal as:

result_image.py

[
  [3],
  [20,9],
  [15,7]
]

It's a translation, but I'll give you a binary tree, so please return it in the list. However, as mentioned above, read it in a zigzag pattern, then ('ω') no .. .. It's like that.

** here ** solves the above problem with BFS, I wondered if I could solve it after studying DFS, so I tried it. For the time being, I passed the description below.

Zig_grouping_DFS.py

class Solution:
    def zigzagLevelOrder(self, root: TreeNode) -> List[List[int]]:
        if not root:
            print("None")
            return

        ans = []
        level = 0
        
        def DFS(node,level):
            if len(ans) == level:
                ans.append([])
            
            ans[level].append(node.val)
            
            if node.left:DFS(node.left,level+1)
            if node.right:DFS(node.right,level+1)
            
            return ans
        
        ZigTree = DFS(root,level)
        
        for i in range(len(ZigTree)):
            if i%2 == 1:
                ZigTree[i].reverse()
                
        return ZigTree
#28ms

Well, the recursive description is simple and good NE !! Let's try deque next time (* ´з`)