解法一
思路
我覺得這是很經典的一道題，運用到了 DFS 和 stack 以及 list 的操作。
We can spearate the problem into three subproblems. Find left boundary, find leaves and find right boundary.
For left boundary, we always check the left child first and then right child if left child is null. If the child is not a leaf, add it into the result list.
For adding leaves, we use dfs to search all leaves from left to right and add them into list. The exit is when we find a leaf -> the node has no child.
For adding right boundary, it is similar to adding left boundary. The only difference is that we check its right child first and then left.
程式碼
class Solution{
public List<Integer> boundaryOfBinaryTree(TreeNode root) {
List<Integer> boundary = new ArrayList<>();
if (root == null) {
return boundary;
}
// No matter left subtree or/and right subtree is null, root is always the first one
boundary.add(root.val);
TreeNode t = root.left;
while (t != null) {
if (!isLeaf(t)) {
boundary.add(t.val);
}
if (t.left != null) {
t =t.left;
} else if (t.right != null) {
t = t.right;
}
}
// Avoid adding root twice is toor itself is a leaf
if (!isLeaf(root)) {
addLeaves(boundary, root);
}
t = root.right;
Stack<Integer> stack = new Stack<>();
while (t != null) {
if (!isLeaf(t)) {
stack.push(t.val);
}
if (t.right != null) {
t = t.right;
} else if (t.left != null) {
t = t.left;
}
}
while (!stack.isEmpty()) {
boundary.add(stack.pop());
}
return boundary;
}
private boolean isLeaf(TreeNode node) {
return node.right == null && node.left == null;
}
private void addLeaves(List<Integer> boundary, TreeNode node) {
if (isLeaf(node)) {
boundary.add(node.val);
} else {
if (node.left != null) {
addLeaves(boundary, node.left);
}
if (node.right != null) {
addLeaves(boundary, node.right);
}
}
}
}複雜度分析
時間複雜度
Worst case O(n) because we traverse each of the node if all nodes are boundary.
空間複雜度
O(n) because we have a stack to store right boundary
解法二
思路
程式碼
複雜度分析
時間複雜度
- 最好情況
- 最壞情況
- 平均情況
空間複雜度