199. Binary Tree Right Side View

Check out the problem description here.

Binary Tree Right Side View - LeetCode

Solution

Approach 1

Iterative using level order traversal.

You can use level order traversal, with traversing from

a)left to right -here the last node in each level is what contribute to the right view

b)right to left -here the f irst node in each level is what contribute to right view

In the code below, we follow left to right level- order traversal.

class Solution:
    def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
        res=[]
        q=[]
        if root:
            q.append([root,0])
        while(q):
            size = len(q)
            for i in range(size):
                front = q.pop(0)
                x=front[1]
                if front[0].left:
                    q.append([front[0].left,x-1])
                if front[0].right:
                    q.append([front[0].right,x+1])
                if(i==size-1):
                    res.append(front[0].val)
        return res

Time:O(N)

Space:O(N)

Approach 2:

Recursive Solution

Traverse the tree in root,right, left fashion.Since rightmost node in each level is the right view for that particular level

class Solution:
    def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
        res=[]
        self.rec(root,0,res)
        return res    
                
    def rec(self,root,level,res):
        if not root:
            return
        if len(res)==level:
            res.append(root.val)
        self.rec(root.right,level+1,res)
        self.rec(root.left,level+1,res)

Time:O(N)

Space:O(h)

h->height of tree (stack space)