Check out the problem description here.

class Solution:
    def binaryTreePaths(self, root: Optional[TreeNode]) -> List[str]:
        res=[]
        ds=[]
        
        self.rec(root,res,ds)
        for i in range(len(res)):
            res[i] = "->".join(res[i])
        return(res)    
           
    def rec(self,root,res,ds):
        if root ==None:
            return
        ds.append(str(root.val))
        
        if root.left ==None and root.right==None:
            res.append(ds.copy())
            ds.pop(-1)
            return
        self.rec(root.left,res,ds)
        self.rec(root.right,res,ds)
        ds.pop(-1)

The above code in a much more neat code is given below

class Solution:
    def binaryTreePaths(self, root: Optional[TreeNode]) -> List[str]:
        res=[]
        if root:
            self.rec(root,"",res)
        return res
    
    def rec(self,root,path,res):
        path+=str(root.val)
        
        if root.right==None and root.left ==None:
            res.append(path)
            return
        if root.left:
            self.rec(root.left,path+"->",res)
        if root.right:
            self.rec(root.right,path+"->",res)

Both solution takes the same time and complexity

TIME :O(n) n= number of nodes

SPACE :O(h) h = height of the tree

Check out the problem description here.

Solution

class Solution:
    def solve(self, A, B):
        ds=[]
        self.rec(A,B,ds)
        return ds
    def rec(self,root,B,ds):
        if root ==None :
            return False
        ds.append(root.val)
        if root.val == B:
            return True
        if self.rec(root.left,B,ds) or self.rec(root.right,B,ds):
            return True
        ds.pop(-1)
        return False

Time :O(N)

Space :O(h) height of the tree

Check the problem description here

Solution

Recursive solution

Recursively check if the mirror nodes are same.

class Solution:
    def isSymmetric(self, root: Optional[TreeNode]) -> bool:
        if root:
            return self.helper(root.left,root.right)
        else:
            return False
    def helper(self,n1,n2):
        if n1==None or n2==None :
            return n1==n2
        if n1.val != n2.val:
            return False
        return (self.helper(n1.left,n2.right) and   self.helper(n1.right,n2.left))

Time :O(N)

Space :O(h) h = height of tree

Check out the problem description here

https://www.codingninjas.com/codestudio/problems/893110?topList=striver-sde-sheet-problems&utm_source=striver&utm_medium=website&leftPanelTab=1

Solution

Traverse level order, and pushing the nodes to queue with the node and its x coordinate(position of imaginary vertical line ) .While popping the node from queue keep updating the imaginary vertical line with latest value popped from the queue and which belong to the corresponding imaginary vertical line .Finally return the dictionary.

def bottomView(root):
    res=[]
    q=[]
    map_str=dict()
    temp=[]
    if root:
        q.append([root,0])
    while(q):
        size = len(q)
        for i in range(size):
            front = q.pop(0)
            x=front[1]
            map_str[x]=front[0].data
            if front[0].left:
                q.append([front[0].left,x-1])
            if front[0].right:
                q.append([front[0].right,x+1])
    for i in sorted(map_str):
        temp.append(map_str[i])
    return temp

Time : O(N)

Space: O(N)

Check out the problem description here.

Solution

Traverse level order, and pushing the nodes to queue with the node and its x coordinate(position of imaginary vertical line ) .While popping the node from queue store the first node value for each imaginary vertical line in to a dictionary .Finally return the dictionary.

def getTopView(root):
    res=[]
    q=[]
    map_str=dict()
    temp=[]
    if root:
        q.append([root,0])
    while(q):
        size = len(q)
        for i in range(size):
            front = q.pop(0)
            x=front[1]
            if x not in map_str:
                map_str[x]=front[0].val
            if front[0].left:
                q.append([front[0].left,x-1])
            if front[0].right:
                q.append([front[0].right,x+1])
    for i in sorted(map_str):
        temp.append(map_str[i])
    return temp

Time : O(N)

Space : O(N)