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].

这道题给了一棵二叉树,让我们以逆时针的顺序来输出树的边界,按顺序分别为左边界,叶结点和右边界。题目中给的例子也很清晰的明白哪些算是边界上的结点。那么最直接的方法就是分别按顺序求出左边界结点,叶结点,和右边界结点。那么如何求的,对于树的操作肯定是用递归最简洁啊,所以可以写分别三个递归函数来分别求左边界结点,叶结点,和右边界结点。首先要处理根结点的情况,当根结点没有左右子结点时,其也是一个叶结点,那么一开始就将其加入结果 res 中,那么再计算叶结点的时候又会再加入一次,这样不对。所以判断如果根结点至少有一个子结点,才提前将其加入结果 res 中。然后再来看求左边界结点的函数,如果当前结点不存在,或者没有子结点,直接返回。否则就把当前结点值加入结果 res 中,然后看如果左子结点存在,就对其调用递归函数,反之如果左子结点不存在,那么对右子结点调用递归函数。而对于求右边界结点的函数就反过来了,如果右子结点存在,就对其调用递归函数,反之如果右子结点不存在,就对左子结点调用递归函数,注意在调用递归函数之后才将结点值加入结果 res,因为是需要按逆时针的顺序输出。最后就来看求叶结点的函数,没什么可说的,就是看没有子结点存在了就加入结果 res,然后对左右子结点分别调用递归即可,参见代码如下:

解法一:

class Solution {
public:
vector<int> boundaryOfBinaryTree(TreeNode* root) {
if (!root) return {};
vector<int> res;
if (root->left || root->right) res.push_back(root->val);
leftBoundary(root->left, res);
leaves(root, res);
rightBoundary(root->right, res);
return res;
}
void leftBoundary(TreeNode* node, vector<int>& res) {
if (!node || (!node->left && !node->right)) return;
res.push_back(node->val);
if (!node->left) leftBoundary(node->right, res);
else leftBoundary(node->left, res);
}
void rightBoundary(TreeNode* node, vector<int>& res) {
if (!node || (!node->left && !node->right)) return;
if (!node->right) rightBoundary(node->left, res);
else rightBoundary(node->right, res);
res.push_back(node->val);
}
void leaves(TreeNode* node, vector<int>& res) {
if (!node) return;
if (!node->left && !node->right) {
res.push_back(node->val);
}
leaves(node->left, res);
leaves(node->right, res);
}
};

下面这种方法把上面三种不同的递归揉合到了一个递归中,并用 bool 型变量来标记当前是求左边界结点还是求右边界结点,同时还有加入叶结点到结果 res 中的功能。如果左边界标记为 true,那么将结点值加入结果 res 中,下面就是调用对左右结点调用递归函数了。根据上面的解题思路可以知道,如果是求左边界结点,优先调用左子结点,当左子结点不存在时再调右子结点,而对于求右边界结点,优先调用右子结点,当右子结点不存在时再调用左子结点。综上考虑,在对左子结点调用递归函数时,左边界标识设为 leftbd && node->left,而对右子结点调用递归的左边界标识设为 leftbd && !node->left,这样左子结点存在就会被优先调用。而右边界结点的情况就正好相反,调用左子结点的右边界标识为 rightbd && !node->right, 调用右子结点的右边界标识为 rightbd && node->right,这样就保证了右子结点存在就会被优先调用,参见代码如下:

解法二:

class Solution {
public:
vector<int> boundaryOfBinaryTree(TreeNode* root) {
if (!root) return {};
vector<int> res{root->val};
helper(root->left, true, false, res);
helper(root->right, false, true, res);
return res;
}
void helper(TreeNode* node, bool leftbd, bool rightbd, vector<int>& res) {
if (!node) return;
if (!node->left && !node->right) {
res.push_back(node->val);
return;
}
if (leftbd) res.push_back(node->val);
helper(node->left, leftbd && node->left, rightbd && !node->right, res);
helper(node->right, leftbd && !node->left, rightbd && node->right, res);
if (rightbd) res.push_back(node->val);
}
};

下面这种解法实际上时解法一的迭代形式,整体思路基本一样,只是没有再用递归的写法,而是均采用 while 的迭代写法,注意在求右边界结点时迭代写法很难直接写出逆时针的顺序,我们可以先反过来保存,最后再调个顺序即可,参见代码如下:

解法三:

class Solution {
public:
vector<int> boundaryOfBinaryTree(TreeNode* root) {
if (!root) return {};
vector<int> res, right;
TreeNode *l = root->left, *r = root->right, *p = root;
if (root->left || root->right) res.push_back(root->val);
while (l && (l->left || l->right)) {
res.push_back(l->val);
if (l->left) l = l->left;
else l = l->right;
}
stack<TreeNode*> st;
while (p || !st.empty()) {
if (p) {
st.push(p);
if (!p->left && !p->right) res.push_back(p->val);
p = p->left;
} else {
p = st.top(); st.pop();
p = p->right;
}
}
while (r && (r->left || r->right)) {
right.push_back(r->val);
if (r->right) r = r->right;
else r = r->left;
}
res.insert(res.end(), right.rbegin(), right.rend());
return res;
}
};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/545

类似题目:

Binary Tree Right Side View

参考资料:

https://leetcode.com/problems/boundary-of-binary-tree/

https://leetcode.com/problems/boundary-of-binary-tree/discuss/101288/Java-Recursive-Solution-Beats-94

https://leetcode.com/problems/boundary-of-binary-tree/discuss/101280/Java(12ms)-left-boundary-left-leaves-right-leaves-right-boundary

https://leetcode.com/problems/boundary-of-binary-tree/discuss/101294/Java-C%2B%2B-Clean-Code-(1-Pass-perorder-postorder-hybrid)

LeetCode All in One 题目讲解汇总(持续更新中...)

[LeetCode] 545. Boundary of Binary Tree 二叉树的边界的更多相关文章

  1. 545. Boundary of Binary Tree二叉树的边界

    [抄题]: Given a binary tree, return the values of its boundary in anti-clockwise direction starting fr ...

  2. [LeetCode] Boundary of Binary Tree 二叉树的边界

    Given a binary tree, return the values of its boundary in anti-clockwise direction starting from roo ...

  3. [LeetCode] Serialize and Deserialize Binary Tree 二叉树的序列化和去序列化

    Serialization is the process of converting a data structure or object into a sequence of bits so tha ...

  4. [LeetCode] 543. Diameter of Binary Tree 二叉树的直径

    Given a binary tree, you need to compute the length of the diameter of the tree. The diameter of a b ...

  5. [LeetCode] Minimum Depth of Binary Tree 二叉树的最小深度

    Given a binary tree, find its minimum depth. The minimum depth is the number of nodes along the shor ...

  6. [LeetCode] Maximum Depth of Binary Tree 二叉树的最大深度

    Given a binary tree, find its maximum depth. The maximum depth is the number of nodes along the long ...

  7. [LeetCode] Maximum Width of Binary Tree 二叉树的最大宽度

    Given a binary tree, write a function to get the maximum width of the given tree. The width of a tre ...

  8. [leetcode]543. Diameter of Binary Tree二叉树直径

    Given a binary tree, you need to compute the length of the diameter of the tree. The diameter of a b ...

  9. [Leetcode] Maximum depth of binary tree二叉树的最大深度

    Given a binary tree, find its maximum depth. The maximum depth is the number of nodes along the long ...

随机推荐

  1. 关于Pragma

    /** This is a introduction of how to use pragma. */ #pragma once /// This is used for include the he ...

  2. background-size:100% 100% 和 background-size:cover的区别简述

    下面我通过给下图背景图添加background-size属性的不同属性值,更直观的显示出100%和cover的区别   下图是添加background-size:100% 100% 后的背景图效果,背 ...

  3. springboot学习源码

    springbootTest 学习源码链接 启动前,需要创建数据库表,修改自己的链接配置 create database test01; use test01; CREATE TABLE catego ...

  4. 【UOJ#74】【UR #6】破解密码

    [UOJ#74][UR #6]破解密码 题面 UOJ 题解 发现这个过程是一个字符串哈希的过程. 把第一位单独拿出来考虑,假设这个串是\(p+S\),旋转后变成了\(S+p\). 其哈希值分别是:\( ...

  5. 【UOJ#60】【UR #5】怎样提高智商

    [UOJ#60][UR #5]怎样提高智商 题面 UOJ 题解 首先猜猜答案是\(4*3^{n-1}\).即前面的选啥都行,后面的搞搞就行了. 而打表(看题解),可以知道答案就是这个,并且每个问题都是 ...

  6. CentOS7系统yum方式安装MySQL5.7

    参考:https://www.cnblogs.com/bigbrotherer/p/7241845.html#top 1.在CentOS中默认安装有MariaDB,这个是MySQL的分支,但为了需要, ...

  7. DEDECMS安全设置怎样做,您就不会说它不安全了

    dedecms是好用,优化也好,就是不安全,个人是这样认为的,今天 闲着没事 整理了一些有助于dedecms网站安全的一些设置,可以说是目前最全的dedecms安全设置! 其一:保持DEDE更新,及时 ...

  8. synchronized和volatile使用

    synchronized和volatile volatile :保证内存可见性,但是不保证原子性: synchronized:同步锁,既能保证内存可见性,又能保证原子性: synchronized实现 ...

  9. java 微信自定义菜单 java微信接口开发 公众平台 SSM redis shiro 多数据源

    A 调用摄像头拍照,自定义裁剪编辑头像,头像图片色度调节B 集成代码生成器 [正反双向](单表.主表.明细表.树形表,快速开发利器)+快速表单构建器 freemaker模版技术 ,0个代码不用写,生成 ...

  10. IntelliJ IDEA 中使用 Lambok (注解无效问题的解决)

    一,概述 Lambok可以说一个能很大提高开发效率的插件,只要在使用注解的方式就能实现很多常用的功能.如常用的@Data能在编译阶段自动生成toString方法,getter方法setter方法等. ...