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A-054● Day 542026.06.30mediumleetcode #199neetcode150

Binary Tree Right Side View

#tree#depth-first-search#breadth-first-search#binary-tree
↗ leetcode #199 · binary-tree-right-side-view
01

Problem

· problem
P.054

Binary Tree Right Side View

↗ leetcode #199

Given the root of a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.

constraints
  • · The number of nodes in the tree is in the range [0, 100]
  • · -100 ≤ Node.val ≤ 100
// paraphrased summary — see source for full text
examples
example 1input → output
[1,2,3,null,5,null,4]
[1,3,4]
example 2input → output
[1,2,3,4,null,null,null,5]
[1,3,4,5]
example 3input → output
[1,null,3]
[1,3]
example 4input → output
[]
[]
02

Pre-solve

· pre-solve
● 1list shown→● 2select→● 3reveal
  • ☐
    Does right side view mean nodes without a right child?
  • ☐
    Should we use BFS or DFS for this problem?
  • ☐
    What should we return for an empty tree?
  • ☐
    Can we only traverse the right subtree of each node?
  • ☐
    Why do we need to know the current level size?
  • ☐
    How do we identify the rightmost node at each level?
  • ☐
    Do we add null nodes to the queue?
check the items you would ask, then press confirm
// session-only state — refresh resets (repeatable practice)
03

Logic Structure

· logic
● 1slots shown→● 2pick per slot→● 3reveal
// pick one code line per slot to assemble the algorithm flow. no typing — just the logic skeleton.
step 1· Initialize result list
○
res = []
○
res = None
○
res = {}
step 2· Initialize queue with root
○
q = collections.deque([root])
○
q = [root]
○
q = collections.deque()
step 3· While loop for each level
○
while q:
○
if q:
○
while len(q) > 1:
step 4· Setup current level│ nested
○
qLen = len(q)
○
qLen = q.maxlen; rightSide = 0
○
qLen = len(q) - 1
step 5· Iterate through current level│ nested
○
for i in range(qLen):
○
for node in q:
○
for i in range(qLen + 1):
step 6· Dequeue and track rightmost│ │ nested
○
node = q.popleft()
○
rightSide = node
q.append(node.left)
q.append(node.right)
○
if node:
                    rightSide = node
step 7· Append rightmost value│ nested
○
if rightSide:
○
res.append(rightSide)
○
res.append(rightSide) if rightSide else pass
pick one option per slot
// format: slot — recursive / DP patterns use ordering / state-first formats. ADR-08 follow-up.
04

Solve · Trace

· solve
solution.py
1
# Definition for a binary tree node.
2
# class TreeNode:
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#     def __init__(self, val=0, left=None, right=None):
4
#         self.val = val
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#         self.left = left
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#         self.right = right
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class Solution:
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    def rightSideView(self, root: TreeNode) -> List[int]:
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        res = []
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        q = collections.deque([root])
11
12
        while q:
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            rightSide = None
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            qLen = len(q)
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            for i in range(qLen):
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                node = q.popleft()
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                if node:
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                    rightSide = node
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                    q.append(node.left)
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                    q.append(node.right)
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            if rightSide:
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                res.append(rightSide.val)
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        return res
mental dry-run cases
// walk each case in your head; expand the worked example below if stuck.
case 1
[1,2,3,null,5,null,4]
→
[1,3,4]
case 2
[1,2,3,4,null,null,null,5]
→
[1,3,4,5]
case 3
[1,null,3]
→
[1,3]
case 4
[]
→
[]
// UI does not walk-through — you do the dry-run mentally. Expand the worked example if stuck.