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11 changes: 11 additions & 0 deletions 0103-binary-tree-zigzag-level-order-traversal/memo.md
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### step1

基本は102. Binary Tree Level Order Traversalと同じで、booleanをもって階層ごとに一つ飛ばしで配列をreverseした。

### step2

変更点なし。

### step3

3回通すまで書き直し。
47 changes: 47 additions & 0 deletions 0103-binary-tree-zigzag-level-order-traversal/step1.cpp
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/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
vector<vector<int>> result;
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result が空であることを利用している点が、ややパズルに感じました。 initializer list を使って以下のように書いたほうがシンプルだと感じました。

if (!root) return {};

vector<vector<int>> result;

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ありがとうございます。initializer listを使った書き方は知りませんでした。

if (!root) return result;
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こちらのコメントをご参照ください。
Shunii85/arai60#17 (comment)


queue<TreeNode*> nodes;
nodes.push(root);

bool toggle = false;
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toggle という変数名に、あまり情報がないように感じました。 right_to_left などはいかがでしょうか?

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そうですね、right_to_leftの方がtrue, falseの意味も分かりやすくていい気がします。

while (!nodes.empty()) {
int size = nodes.size();
vector<int> level;

for (int i = 0; i < size; i++) {
TreeNode* node = nodes.front();
nodes.pop();

level.push_back(node->val);

if (node->left) nodes.push(node->left);
if (node->right) nodes.push(node->right);
}

if (toggle) {
reverse(level.begin(), level.end());
}

toggle = !toggle;

result.push_back(level);
}

return result;
}
};
48 changes: 48 additions & 0 deletions 0103-binary-tree-zigzag-level-order-traversal/step2.cpp
Original file line number Diff line number Diff line change
@@ -0,0 +1,48 @@
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
vector<vector<int>> result;
if (!root) {
return result;
}

queue<TreeNode*> nodes;
nodes.push(root);

bool toggle = false;
while (!nodes.empty()) {
int size = nodes.size();
vector<int> level;

for (int i = 0; i < size; i++) {
TreeNode* node = nodes.front();
nodes.pop();

level.push_back(node->val);
if (node->left) {
nodes.push(node->left);
}
if (node->right) {
nodes.push(node->right);
}
}
if (toggle) {
reverse(level.begin(), level.end());
}
toggle = !toggle;
result.push_back(level);
}
return result;
}
};
48 changes: 48 additions & 0 deletions 0103-binary-tree-zigzag-level-order-traversal/step3.cpp
Original file line number Diff line number Diff line change
@@ -0,0 +1,48 @@
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
vector<vector<int>> result;
if (!root) {
return result;
}

queue<TreeNode*> nodes;
nodes.push(root);

bool toggle = false;
while (!nodes.empty()) {
int size = nodes.size();
vector<int> level;

for (int i = 0; i < size; i++) {
TreeNode* node = nodes.front();
nodes.pop();

level.push_back(node->val);
if (node->left) {
nodes.push(node->left);
}
if (node->right) {
nodes.push(node->right);
}
}
if (toggle) {
reverse(level.begin(), level.end());
}
toggle = !toggle;
result.push_back(level);
}
return result;
}
};