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Binary Tree Right Side View

The right view needs exactly one node per depth: the first one DFS reaches when it always visits the right child before the left. A node fills a depth only the first time DFS arrives there — any later arrival at the same depth (from a shorter right branch or a deeper left branch) is invisible from the right.

Details

Animated on: Given the root of a binary tree, return the values of the nodes visible from the right side, ordered from top to bottom.

Problem:Binary Tree Right Side ViewMedium

Pattern:Tree BFS and Level Order

Binary Trees

step 1 / 17

4
2
1
3
5

Call rightSideView(1). result starts empty; DFS explores right children before left so the first node to reach each new depth is always the rightmost one.

Pseudocode
1FUNCTION rightSideView(root):
2  result <- []
3  DFS(root, 0, result)
4  RETURN result
5FUNCTION DFS(node, depth, result):
6  IF node = null: RETURN
7  IF depth = LENGTH(result):
8    APPEND node.val TO result
9  DFS(node.right, depth + 1, result)
10  DFS(node.left, depth + 1, result)

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Every step in words

Show all 17 steps
  1. Call rightSideView(1). result starts empty; DFS explores right children before left so the first node to reach each new depth is always the rightmost one.
  2. Enter DFS(1, depth=0). Call stack: DFS(1, depth=0).
  3. depth (0) equals result.length (0) — first arrival at this level, so 1 is the rightmost node seen so far. result = [1].
  4. Enter DFS(3, depth=1). Call stack: DFS(1, depth=0) -> DFS(3, depth=1).
  5. depth (1) equals result.length (1) — first arrival at this level, so 3 is the rightmost node seen so far. result = [1, 3].
  6. Enter DFS(5, depth=2). Call stack: DFS(1, depth=0) -> DFS(3, depth=1) -> DFS(5, depth=2).
  7. depth (2) equals result.length (2) — first arrival at this level, so 5 is the rightmost node seen so far. result = [1, 3, 5].
  8. 5 has no children — both recursive calls hit the null base case immediately, so DFS(5, depth=2) returns and pops.
  9. Both subtrees of 3 are done. DFS(3, depth=1) returns and pops. result so far: [1, 3, 5].
  10. Enter DFS(2, depth=1). Call stack: DFS(1, depth=0) -> DFS(2, depth=1).
  11. depth (1) != result.length (3) — a node further right already filled this level, so 2 stays hidden. result unchanged: [1, 3, 5].
  12. Enter DFS(4, depth=2). Call stack: DFS(1, depth=0) -> DFS(2, depth=1) -> DFS(4, depth=2).
  13. depth (2) != result.length (3) — a node further right already filled this level, so 4 stays hidden. result unchanged: [1, 3, 5].
  14. 4 has no children — both recursive calls hit the null base case immediately, so DFS(4, depth=2) returns and pops.
  15. Both subtrees of 2 are done. DFS(2, depth=1) returns and pops. result so far: [1, 3, 5].
  16. Both subtrees of 1 are done. DFS(1, depth=0) returns and pops. result so far: [1, 3, 5].
  17. DFS finished. result = [1, 3, 5] — the highlighted path shows the rightmost node at every depth.

Now solve it yourself

Watching is the first half. The second is writing it without the animation.

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