(M)BFS:103. Binary Tree Zigzag Level Order Traversal

这道题意思是按照“之”字形遍历一棵树。

我的做法是dfs把每一行按照正常的顺序存下来，最后把奇数行倒置。

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    vector<vector<int>> zigzagLevelOrder(TreeNode* root) {        vector<vector<int>> res;        findorder(root, res, 1);        for(int i = 0; i < res.size(); ++i)        {            if(i % 2 != 0)                reverse(res[i].begin(), res[i].end());        }        return res;    }    void findorder(TreeNode* root, vector<vector<int>>& res, int height)    {        if(root == NULL)    return;        if(res.size() < height)        {            res.push_back({root->val});        }        else        {            res[height - 1].push_back(root->val);        }        findorder(root->left, res, height+1);        findorder(root->right, res, height+1);    }};

但是这道题归在BFS里，应该想到这种变换顺序的遍历可以用栈来实现：

根据其特点我们用到栈的后进先出的特点，这道题我们维护两个栈，相邻两行分别存到两个栈中，进栈的顺序也不相同，一个栈是先进左子结点然后右子节点，另一个栈是先进右子节点然后左子结点，这样出栈的顺序就是我们想要的之字形了。

class Solution {public:    vector<vector<int> > zigzagLevelOrder(TreeNode *root) {        vector<vector<int> >res;        if (!root) return res;        stack<TreeNode*> s1;        stack<TreeNode*> s2;        s1.push(root);        vector<int> out;        while (!s1.empty() || !s2.empty()) {            while (!s1.empty()) {                TreeNode *cur = s1.top();                s1.pop();                out.push_back(cur->val);                if (cur->left) s2.push(cur->left);                if (cur->right) s2.push(cur->right);            }             if (!out.empty()) res.push_back(out);            out.clear();            while (!s2.empty()) {                TreeNode *cur = s2.top();                s2.pop();                out.push_back(cur->val);                if (cur->right) s1.push(cur->right);                if (cur->left) s1.push(cur->left);            }            if (!out.empty()) res.push_back(out);            out.clear();        }        return res;    }};