使用 dfs 遍历二叉树，在给定点停止（在 Python 中）

Question

I'm learning about some basic computer science concepts. As a demo, I'm creating a script in Python that will perform various functions on a binary tree. I've been able to successfully program most of these functions, except for one specific one.

For a binary tree created from an array of integers, I would like to do a depth first search for an integer, and return the path that was taken through the tree up until that integer was found as an array (with the last number in that array being the number that was being searched for). As soon as the first match for that integer is found, I want to stop traversing the tree.

For example, for an inorder dfs of array [3,2,4,1,4,6,8,5] for integer 4, it should return [1,2,3,4]

For integer 5 it should return [1,2,3,4,4,5] etc.

Here is my code:

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

    def getValue(self):
        return self.value

def buildTree(array):
    print("building tree....")
    root=Node(array[0])
    del(array[0])
    for a in array:
        insert(root,Node(a))
    print("building complete")
    return root

def insert(root,node): 
    if root is None: 
        root = node 
    else: 
        if root.value < node.value: 
            if root.right is None: 
                root.right = node 
            else: 
                insert(root.right, node) 
        else: 
            if root.left is None: 
                root.left = node 
            else: 
                insert(root.left, node) 

def depthFirstSearch(root,target,results,subSearch):
#0:preorder
#1:inorder
#2:postorder
    if root!=None:

        if subSearch==0:
            results.append(root.getValue())
            if root.getValue()==target:
                return results

        depthFirstSearch(root.left,target,results,subSearch)

        if subSearch==1:
            results.append(root.getValue())
            if root.getValue()==target:
                return results

        depthFirstSearch(root.right,target,results,subSearch)

        if subSearch==2:
            results.append(root.getValue())
            if root.getValue()==target:
                return results

    return results

if __name__ == '__main__':
    #stuff that gets our arguments
    #...
    array=[3,2,4,1,4,6,8,5] #we would actually get this as an argument, but using this for example array
    target=4 #example target
    subSearch=1 #using inorder traversal for this example

    root=buildTree(array)
    results=[]
    results=depthFirstSearch(root,target,results,subSearch) 
    print(results) #expected:[1,2,3,4]

Answer 1

Okay this is easy, just use an Additional variable flag and then your function becomes

def depthFirstSearch(root,target,results,subSearch, flag = 0):
#0:preorder
#1:inorder
#2:postorder
    if root!=None:

        if subSearch==0:
            results.append(root.getValue())
            if root.getValue()==target:
                return results, 1

        results, flag =depthFirstSearch(root.left,target,results,subSearch)
        if flag == 1:
            return results, flag
        if subSearch==1:
            results.append(root.getValue())
            if root.getValue()==target:
                return results, 1

        results, flag = depthFirstSearch(root.right,target,results,subSearch)
        if flag == 1:
            return results, flag

        if subSearch==2:
            results.append(root.getValue())
            if root.getValue()==target:
                return results, 1

    return results, flag

Here Flag changes to one and that is propagated as the function stack shrinks, keeping them after each recursive calls takes care of this.

Also in main function, the function call becomes

results, _=depthFirstSearch(root,target,results,subSearch)

Since flag = 0 is present in function definition you just need to discard the second variable, you can even use that to check if the element was found in the tree rather than just printing the whole tree if element is not present.

If you have any doubts or concerns, comment below.