(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;
    }
};