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feat: add solutions to lc problem: No.0545. Boundary of Binary Tree
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‎README.md

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@@ -125,6 +125,7 @@
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- [二叉搜索树的最近公共祖先](/solution/0200-0299/0236.Lowest%20Common%20Ancestor%20of%20a%20Binary%20Tree/README.md)
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- [将二叉搜索树转换为单链表](/lcci/17.12.BiNode/README.md)
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- [将二叉搜索树转化为排序的双向链表](/solution/0400-0499/0426.Convert%20Binary%20Search%20Tree%20to%20Sorted%20Doubly%20Linked%20List/README.md)
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- [二叉树的边界](/solution/0500-0599/0545.Boundary%20of%20Binary%20Tree/README.md)
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### 数学
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‎README_EN.md

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@@ -121,6 +121,7 @@ Complete solutions to [LeetCode](https://leetcode-cn.com/problemset/all/), [LCOF
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- [Lowest Common Ancestor of a Binary Search Tree](/solution/0200-0299/0235.Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree/README_EN.md)
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- [BiNode](/lcci/17.12.BiNode/README_EN.md)
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- [Convert Binary Search Tree to Sorted Doubly Linked List](/solution/0400-0499/0426.Convert%20Binary%20Search%20Tree%20to%20Sorted%20Doubly%20Linked%20List/README_EN.md)
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- [Boundary of Binary Tree](/solution/0500-0599/0545.Boundary%20of%20Binary%20Tree/README_EN.md)
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### Math
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‎solution/0100-0199/0199.Binary Tree Right Side View/README.md

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Original file line numberDiff line numberDiff line change
@@ -45,13 +45,13 @@ class Solution:
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def rightSideView(self, root: TreeNode) -> List[int]:
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if not root:
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return []
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q = [root]
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q = collections.deque([root])
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res = []
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while q:
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size = len(q)
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res.append(q[0].val)
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for _ in range(size):
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node = q.pop(0)
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node = q.popleft()
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if node.right:
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q.append(node.right)
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if node.left:

‎solution/0100-0199/0199.Binary Tree Right Side View/README_EN.md

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Original file line numberDiff line numberDiff line change
@@ -54,13 +54,13 @@ class Solution:
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def rightSideView(self, root: TreeNode) -> List[int]:
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if not root:
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return []
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q = [root]
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q = collections.deque([root])
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res = []
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while q:
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size = len(q)
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res.append(q[0].val)
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for _ in range(size):
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node = q.pop(0)
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node = q.popleft()
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if node.right:
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q.append(node.right)
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if node.left:

‎solution/0100-0199/0199.Binary Tree Right Side View/Solution.py

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Original file line numberDiff line numberDiff line change
@@ -8,13 +8,13 @@ class Solution:
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def rightSideView(self, root: TreeNode) -> List[int]:
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if not root:
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return []
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q = [root]
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q = collections.deque([root])
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res = []
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while q:
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size = len(q)
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res.append(q[0].val)
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for _ in range(size):
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node = q.pop(0)
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node = q.popleft()
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if node.right:
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q.append(node.right)
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if node.left:

‎solution/0500-0599/0545.Boundary of Binary Tree/README.md

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@@ -56,18 +56,70 @@
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<li><code>-1000 <= Node.val <= 1000</code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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分别求根节点、左边界、叶子节点、右边界，依次放入结果数组 res 中。
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注意，求右边界的时候，需要逆序结果，这时可以用栈实现。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# 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 boundaryOfBinaryTree(self, root: TreeNode) -> List[int]:
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self.res = []
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if not root:
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return self.res
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# root
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if not self.is_leaf(root):
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self.res.append(root.val)
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# left boundary
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t = root.left
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while t:
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if not self.is_leaf(t):
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self.res.append(t.val)
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t = t.left if t.left else t.right
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# leaves
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self.add_leaves(root)
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# right boundary(reverse order)
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s = []
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t = root.right
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while t:
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if not self.is_leaf(t):
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s.append(t.val)
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t = t.right if t.right else t.left
106+
while s:
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self.res.append(s.pop())
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# output
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return self.res
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def add_leaves(self, root):
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if self.is_leaf(root):
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self.res.append(root.val)
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return
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if root.left:
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self.add_leaves(root.left)
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if root.right:
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self.add_leaves(root.right)
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def is_leaf(self, node) -> bool:
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return node and node.left is None and node.right is None
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```
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@@ -76,7 +128,81 @@
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
79-
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/**
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* Definition for a binary tree node.
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* public class TreeNode {
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* int val;
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* TreeNode left;
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* TreeNode right;
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* TreeNode() {}
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* TreeNode(int val) { this.val = val; }
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* TreeNode(int val, TreeNode left, TreeNode right) {
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* this.val = val;
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* this.left = left;
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* this.right = right;
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* }
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* }
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*/
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class Solution {
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private List<Integer> res;
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public List<Integer> boundaryOfBinaryTree(TreeNode root) {
150+
if (root == null) {
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return Collections.emptyList();
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}
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res = new ArrayList<>();
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// root
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if (!isLeaf(root)) {
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res.add(root.val);
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}
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// left boundary
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TreeNode t = root.left;
162+
while (t != null) {
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if (!isLeaf(t)) {
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res.add(t.val);
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}
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t = t.left == null ? t.right : t.left;
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}
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// leaves
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addLeaves(root);
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// right boundary(reverse order)
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Deque<Integer> s = new ArrayDeque<>();
174+
t = root.right;
175+
while (t != null) {
176+
if (!isLeaf(t)) {
177+
s.offer(t.val);
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}
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t = t.right == null ? t.left : t.right;
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}
181+
while (!s.isEmpty()) {
182+
res.add(s.pollLast());
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}
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// output
186+
return res;
187+
}
188+
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private void addLeaves(TreeNode root) {
190+
if (isLeaf(root)) {
191+
res.add(root.val);
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return;
193+
}
194+
if (root.left != null) {
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addLeaves(root.left);
196+
}
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if (root.right != null) {
198+
addLeaves(root.right);
199+
}
200+
}
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202+
private boolean isLeaf(TreeNode node) {
203+
return node != null && node.left == null && node.right == null;
204+
}
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}
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```
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### **...**

‎solution/0500-0599/0545.Boundary of Binary Tree/README_EN.md

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@@ -57,21 +57,143 @@ Concatenating everything results in [1] + [2] + [4,7,8,9,10] + [6,3] = [1,2,4,7,
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<li><code>-1000 &lt;= Node.val &lt;= 1000</code></li>
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</ul>
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## Solutions
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<!-- tabs:start -->
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### **Python3**
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```python
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# Definition for a binary tree node.
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# class TreeNode:
69+
# def __init__(self, val=0, left=None, right=None):
70+
# self.val = val
71+
# self.left = left
72+
# self.right = right
73+
class Solution:
74+
def boundaryOfBinaryTree(self, root: TreeNode) -> List[int]:
75+
self.res = []
76+
if not root:
77+
return self.res
78+
# root
79+
if not self.is_leaf(root):
80+
self.res.append(root.val)
81+
82+
# left boundary
83+
t = root.left
84+
while t:
85+
if not self.is_leaf(t):
86+
self.res.append(t.val)
87+
t = t.left if t.left else t.right
88+
89+
# leaves
90+
self.add_leaves(root)
91+
92+
# right boundary(reverse order)
93+
s = []
94+
t = root.right
95+
while t:
96+
if not self.is_leaf(t):
97+
s.append(t.val)
98+
t = t.right if t.right else t.left
99+
while s:
100+
self.res.append(s.pop())
101+
102+
# output
103+
return self.res
104+
105+
def add_leaves(self, root):
106+
if self.is_leaf(root):
107+
self.res.append(root.val)
108+
return
109+
if root.left:
110+
self.add_leaves(root.left)
111+
if root.right:
112+
self.add_leaves(root.right)
113+
114+
def is_leaf(self, node) -> bool:
115+
return node and node.left is None and node.right is None
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69117
```
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### **Java**
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73121
```java
74-
122+
/**
123+
* Definition for a binary tree node.
124+
* public class TreeNode {
125+
* int val;
126+
* TreeNode left;
127+
* TreeNode right;
128+
* TreeNode() {}
129+
* TreeNode(int val) { this.val = val; }
130+
* TreeNode(int val, TreeNode left, TreeNode right) {
131+
* this.val = val;
132+
* this.left = left;
133+
* this.right = right;
134+
* }
135+
* }
136+
*/
137+
class Solution {
138+
private List<Integer> res;
139+
140+
public List<Integer> boundaryOfBinaryTree(TreeNode root) {
141+
if (root == null) {
142+
return Collections.emptyList();
143+
}
144+
res = new ArrayList<>();
145+
146+
// root
147+
if (!isLeaf(root)) {
148+
res.add(root.val);
149+
}
150+
151+
// left boundary
152+
TreeNode t = root.left;
153+
while (t != null) {
154+
if (!isLeaf(t)) {
155+
res.add(t.val);
156+
}
157+
t = t.left == null ? t.right : t.left;
158+
}
159+
160+
// leaves
161+
addLeaves(root);
162+
163+
// right boundary(reverse order)
164+
Deque<Integer> s = new ArrayDeque<>();
165+
t = root.right;
166+
while (t != null) {
167+
if (!isLeaf(t)) {
168+
s.offer(t.val);
169+
}
170+
t = t.right == null ? t.left : t.right;
171+
}
172+
while (!s.isEmpty()) {
173+
res.add(s.pollLast());
174+
}
175+
176+
// output
177+
return res;
178+
}
179+
180+
private void addLeaves(TreeNode root) {
181+
if (isLeaf(root)) {
182+
res.add(root.val);
183+
return;
184+
}
185+
if (root.left != null) {
186+
addLeaves(root.left);
187+
}
188+
if (root.right != null) {
189+
addLeaves(root.right);
190+
}
191+
}
192+
193+
private boolean isLeaf(TreeNode node) {
194+
return node != null && node.left == null && node.right == null;
195+
}
196+
}
75197
```
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### **...**

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