545. Boundary of Binary Tree
Given a binary tree, return the values of its boundary in anti-clockwise direction starting from root.
Boundary includes left boundary, leaves, and right boundary in order without duplicate nodes. (The values
of the nodes may still be duplicates.)
Left boundary is defined as the path from root to the left-most node. Right boundary is defined as the
path from root to the right-most node. If the root doesn't have left subtree or right subtree, then the
root itself is left boundary or right boundary. Note this definition only applies to the input binary
tree, and not applies to any subtrees.
The left-most node is defined as a leaf node you could reach when you always firstly travel to the left
subtree if exists. If not, travel to the right subtree. Repeat until you reach a leaf node.
The right-most node is also defined by the same way with left and right exchanged.
Example 1
Input:
1
\
2
/ \
3 4
Ouput:
[1, 3, 4, 2]
Explanation:
The root doesn't have left subtree, so the root itself is left boundary.
The leaves are node 3 and 4.
The right boundary are node 1,2,4. Note the anti-clockwise direction means you should output reversed
right boundary.
So order them in anti-clockwise without duplicates and we have [1,3,4,2].
Example 2
Input:
____1_____
/ \
2 3
/ \ /
4 5 6
/ \ / \
7 8 9 10
Ouput:
[1,2,4,7,8,9,10,6,3]
Explanation:
The left boundary are node 1,2,4. (4 is the left-most node according to definition)
The leaves are node 4,7,8,9,10.
The right boundary are node 1,3,6,10. (10 is the right-most node).
So order them in anti-clockwise without duplicate nodes we have [1,2,4,7,8,9,10,6,3].
- A fast coded answer
Searching for leftmost and rightmost and leaves are overlapped, so i’ll try to optimize the solution. Hope to solve it with one pass inorder traverse
A State Machine solution is also feasible, here are four states: on left boundary, right boundary, leaves, internal nodes.
class Solution {
public List<Integer> boundaryOfBinaryTree(TreeNode root) {
ArrayList<Integer> result = new ArrayList<>();
if(root == null) return result;
HashSet<TreeNode> added = new HashSet<>();
// left boundary
TreeNode curr = root;
added.add(curr);
result.add(curr.val);
if(curr.left != null) {
curr = curr.left;
while(curr != null) {
added.add(curr);
result.add(curr.val);
if(curr.left != null) curr = curr.left;
else curr = curr.right;
}
}
// right boundary
ArrayList<Integer> temp = new ArrayList<>();
curr = root;
if(curr.right != null) {
curr = curr.right;
while(curr != null) {
if(!added.contains(curr)) {
temp.add(curr.val);
added.add(curr);
}
curr = curr.right != null ? curr.right : curr.left;
}
}
// collect leaves
collect(root, added, result);
for(int i = temp.size()-1; i >= 0; i--) result.add(temp.get(i));
return result;
}
public void collect(TreeNode root, HashSet<TreeNode> added, ArrayList<Integer> result) {
if(root == null) return;
if(root.left == null && root.right == null && !added.contains(root)) {
result.add(root.val);
added.add(root);
}
collect(root.left, added, result);
collect(root.right, added, result);
}
}