Code & Func

第97天。

今天的题目是Binary Tree Zigzag Level Order Traversal:

Given a binary tree, return the zigzag level order traversal of its nodes’ values. (ie, from left to right, then right to left for the next level and alternate between).

For example: Given binary tree [3,9,20,null,null,15,7],

return its zigzag level order traversal as:

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[
2
  [3],
3
  [20,9],
4
  [15,7]
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]

首先想到的是用层次遍历的方式来实现。

简单的层次遍历：

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void levelTra(TreeNode *root) {
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    if (root == nullptr) return ;
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    queue<TreeNode *> q;
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    q.push(root);
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    while(!q.empty()) {
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        root = q.front(); q.pop();
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        cout << root->val;
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        if (root->left) q.push(root->left);
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        if (root->right) q.push(root->right);
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    }
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}

但是上面的方法是没法区分层数的，我们通过nullptr来表示换行：

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void levelTra(TreeNode *root) {
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    if (root == nullptr) return ;
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    queue<TreeNode *> q;
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    q.push(root);
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    q.push(nullptr);
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    while(true) {
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        root = q.top(); q.pop();
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        if (root == nullptr) {
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            cout << "new level" << endl;
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            if (q.empty()) break;
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            q.push(nullptr);
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        }
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        cout << root->val << endl;
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        if (root->left) q.push(root->left);
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        if (root->right) q.push(root->right);
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    }
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}

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vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
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    vector<vector<int> > ret;
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    vector<int> tmp;
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    if (root == nullptr) return ret;
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    //level tra
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    bool flag = true; //判断遍历方向
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    queue<TreeNode *> q;
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    q.push(root);
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    q.push(nullptr);
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    while(true) {
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        root = q.front(); q.pop();
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        if (root == nullptr)  {
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            if (flag) ret.push_back(tmp);
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            else ret.push_back(vector<int>(tmp.rbegin(), tmp.rend()));
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            tmp.clear();
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            flag = !flag;
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            if (q.empty()) break;
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            q.push(nullptr);
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            continue;
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        }
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        tmp.push_back(root->val);
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        if (root->left) q.push(root->left);
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        if (root->right) q.push(root->right);
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    }
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    return ret;
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}

然后是dicuss中的方法，简单的来说就是通过深度优先遍历来生成获取层次遍历的每层的数组（好像之前写过？），然后就会比前面用queue的方法快。

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void travel(TreeNode *root, vector<vector<int> > &ret, int level) {
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    if (root == nullptr) return ;
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    if (level >= ret.size()) ret.push_back(vector<int>());
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    ret[level].push_back(root->val);
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    travel(root->left, ret, level + 1);
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    travel(root->right, ret, level + 1);
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}
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vector<vector<int> > zigzagLevelOrder(TreeNode *root) {
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    vector<vector<int>> ret;
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    travel(root, ret, 0);
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    for(int i = 0;i < ret.size();i++) {
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        if (i % 2) reverse(ret[i].begin(), ret[i].end());
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    }
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    return ret;
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}

Update as 2020-03-28

最近在总结 Stack Tag 的算法，然后发现这道题可以用双栈来解，和前面队列的做法有点类似，某种意义上也是在模拟层次遍历，但是因为栈后进先出的特性，所以很直接的实现了逆序的操作，不需要额外做reverse。

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vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
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    vector<vector<int> > res;
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    int level = 0;
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    stack<TreeNode*> st_even, st_odd;
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    if (root) st_even.push(root);
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    while(!st_even.empty() || !st_odd.empty()) {
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        stack<TreeNode*> &st1 = level % 2 == 0 ? st_even : st_odd;
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        stack<TreeNode*> &st2 = level % 2 == 0 ? st_odd : st_even;
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        vector<int> temp;
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        for(int i = 0, size = st1.size(); i < size; i++) {
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            root = st1.top(); st1.pop();
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            temp.push_back(root->val);
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            TreeNode *left = root->left, *right = root->right;
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            if (level % 2) swap(left, right);
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            if (left) st2.push(left);
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            if (right) st2.push(right);
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        }
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        //cout << st1.size() << " " << st2.size() << endl;
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        res.push_back(temp);
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        level++;
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    }
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    return res;
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}