［抄题］：
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.
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].
[暴力解法]：
时间分析：
空间分析：
[优化后]：
时间分析：
空间分析：
［奇葩输出条件］：
［奇葩corner case］：
［思维问题］：
想不到“边界”应该怎么控制：用boolean变量表示方向l/r
主函数中设置t f，强制性设置好条件往左右扩展。dfs函数中设置== null, 有条件才往左右扩展，能走多深走多深。
［英文数据结构或算法，为什么不用别的数据结构或算法］：
［一句话思路］：
［输入量］：空： 正常情况：特大：特小：程序里处理到的特殊情况：异常情况（不合法不合理的输入）：
［画图］：
［一刷］：
- dfs一直是对当前节点操作的，不是node.left/node.right
- 因为dfs的函数是左右分开的，不存在一了百了的情况。所以主函数也要一个点先进去，然后再进行dc。
［二刷］：
［三刷］：
［四刷］：
［五刷］：
[五分钟肉眼debug的结果]：
［总结］：
想不到“边界”应该怎么控制：用boolean变量表示方向l/r
［复杂度］：Time complexity: O(n) Space complexity: O(n)
［算法思想：迭代/递归/分治/贪心］：
［关键模板化代码］：
solution要用来新建answer对象。然后TreeNode root需要在主函数中再声明一遍。
class MyCode { public static void main (String[] args) { Solution answer = new Solution(); answer.root = new TreeNode(1); answer.root.right = new TreeNode(2); answer.root.right.left = new TreeNode(3); answer.root.right.right = new TreeNode(4); List<Integer> result = answer.boundaryOfBinaryTree(answer.root); for (int i = 0; i < result.size(); i++) System.out.println("result.get(i) = " + result.get(i)); }
［其他解法］：
［Follow Up］：
［LC给出的题目变变变］：
[代码风格] ：
[是否头一次写此类driver funcion的代码] ：
[潜台词] ：
// package whatever; // don't place package name! import java.io.*; import java.util.*; import java.lang.*; class TreeNode { int val; TreeNode left, right; //parameter is another item TreeNode(int item) { val = item; left = right = null; } } class Solution { TreeNode root; public List<Integer> boundaryOfBinaryTree(TreeNode root) { //initialization List<Integer> result = new ArrayList<Integer>(); //corner case if (root == null) return result; //dfs in left and right result.add(root.val); dfs(root.left, true, false, result); dfs(root.right, false, true, result); //return return result; } public void dfs(TreeNode root, boolean lb, boolean rb, List<Integer> result) { //exit case if (root == null) return ; //add the left root if (lb) result.add(root.val); //add the mid root if (!lb && !rb && root.left == null && root.right == null) result.add(root.val); //dfs in left and right dfs(root.left, lb, rb && root.left == null, result); dfs(root.right, lb && root.left == null, rb, result); //add the right root if (rb) result.add(root.val); } } class MyCode { public static void main (String[] args) { Solution answer = new Solution(); answer.root = new TreeNode(1); answer.root.right = new TreeNode(2); answer.root.right.left = new TreeNode(3); answer.root.right.right = new TreeNode(4); List<Integer> result = answer.boundaryOfBinaryTree(answer.root); for (int i = 0; i < result.size(); i++) System.out.println("result.get(i) = " + result.get(i)); } }