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Commit e548c2a

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raof01azl397985856
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feat: azl397985856#103, azl397985856#129: add C++ implementation (azl397985856#63)
* Update 103.binary-tree-zigzag-level-order-traversal.md * azl397985856#129: add C++ implementation 写描述比写代码难。。。
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‎problems/103.binary-tree-zigzag-level-order-traversal.md‎

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@@ -44,6 +44,10 @@ return its zigzag level order traversal as:
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## 代码
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* 语言支持:JS,C++
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JavaScript Code:
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```js
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/*
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* @lc app=leetcode id=103 lang=javascript
@@ -132,6 +136,77 @@ var zigzagLevelOrder = function(root) {
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};
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```
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C++ Code:
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```C++
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
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* };
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*/
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class Solution {
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public:
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vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
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auto ret = vector<vector<int>>();
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if (root == nullptr) return ret;
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auto queue = vector<const TreeNode*>{root};
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auto isOdd = true;
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while (!queue.empty()) {
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auto sz = queue.size();
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auto level = vector<int>();
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for (auto i = 0; i < sz; ++i) {
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auto n = queue.front();
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queue.erase(queue.begin());
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if (isOdd) level.push_back(n->val);
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else level.insert(level.begin(), n->val);
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if (n->left != nullptr) queue.push_back(n->left);
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if (n->right != nullptr) queue.push_back(n->right);
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}
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isOdd = !isOdd;
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ret.push_back(level);
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}
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return ret;
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}
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};
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```
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## 拓展
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由于二叉树是递归结构,因此,可以采用递归的方式来处理。在递归时需要保留当前的层次信息(从0开始),作为参数传递给下一次递归调用。
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### 描述
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1. 当前层次为偶数时,将当前节点放到当前层的结果数组尾部
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2. 当前层次为奇数时,将当前节点放到当前层的结果数组头部
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3. 递归对左子树进行之字形遍历,层数参数为当前层数+1
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4. 递归对右子树进行之字形遍历,层数参数为当前层数+1
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### C++实现
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```C++
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class Solution {
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public:
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vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
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auto ret = vector<vector<int>>();
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zigzagLevelOrder(root, 0, ret);
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return ret;
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}
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private:
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void zigzagLevelOrder(const TreeNode* root, int level, vector<vector<int>>& ret) {
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if (root == nullptr || level < 0) return;
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if (ret.size() <= level) {
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ret.push_back(vector<int>());
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}
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if (level % 2 == 0) ret[level].push_back(root->val);
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else ret[level].insert(ret[level].begin(), root->val);
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zigzagLevelOrder(root->left, level + 1, ret);
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zigzagLevelOrder(root->right, level + 1, ret);
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}
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};
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```
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## 相关题目
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- [102.binary-tree-level-order-traversal](./102.binary-tree-level-order-traversal.md)
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- [104.maximum-depth-of-binary-tree](./104.maximum-depth-of-binary-tree.md)

‎problems/129.sum-root-to-leaf-numbers.md‎

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@@ -64,6 +64,10 @@ Therefore, sum = 495 +たす 491 +たす 40 =わ 1026.
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## 代码
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* 语言支持:JS,C++
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JavaScipt Code:
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```js
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/*
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* @lc app=leetcode id=129 lang=javascript
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};
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```
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C++ Code:
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```C++
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
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* };
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*/
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class Solution {
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public:
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int sumNumbers(TreeNode* root) {
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return helper(root, 0);
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}
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private:
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int helper(const TreeNode* root, int val) {
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if (root == nullptr) return 0;
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auto ret = root->val + val * 10;
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if (root->left == nullptr && root->right == nullptr)
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return ret;
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auto l = helper(root->left, ret);
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auto r = helper(root->right, ret);
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return l + r;
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}
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};
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```
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## 拓展
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通常来说,可以利用队列、栈等数据结构将递归算法转为递推算法。
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### 描述
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使用两个队列:
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1. 当前和队列:保存上一层每个结点的当前和(比如49和40)
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2. 结点队列:保存当前层所有的非空结点
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每次循环按层处理结点队列。处理步骤:
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1. 从结点队列取出一个结点
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2. 从当前和队列将上一层对应的当前和取出来
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3. 若左子树非空,则将该值乘以10加上左子树的值,并添加到当前和队列中
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4. 若右子树非空,则将该值乘以10加上右子树的值,并添加到当前和队列中
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5. 若左右子树均为空时,将该节点的当前和加到返回值中
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### C++实现
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```C++
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class Solution {
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public:
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int sumNumbers(TreeNode* root) {
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if (root == nullptr) return 0;
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auto ret = 0;
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auto runningSum = vector<int>{root->val};
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auto queue = vector<const TreeNode*>{root};
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while (!queue.empty()) {
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auto sz = queue.size();
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for (auto i = 0; i < sz; ++i) {
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auto n = queue.front();
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queue.erase(queue.begin());
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auto tmp = runningSum.front();
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runningSum.erase(runningSum.begin());
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if (n->left != nullptr) {
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runningSum.push_back(tmp * 10 + n->left->val);
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queue.push_back(n->left);
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}
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if (n->right != nullptr) {
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runningSum.push_back(tmp * 10 + n->right->val);
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queue.push_back(n->right);
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}
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if (n->left == nullptr && n->right == nullptr) {
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ret += tmp;
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}
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}
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}
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return ret;
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}
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};
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```
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## 相关题目
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- [sum-of-root-to-leaf-binary-numbers](https://leetcode.com/problems/sum-of-root-to-leaf-binary-numbers/)

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