LeetCode :: Binary Tree Zigzag Level Order Traversal [tree, BFS]

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,#,#,15,7},

    3   / \  9  20    /  \   15   7

return its zigzag level order traversal as:

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

题目的意思很直白，层序遍历整个树，但是第一层正序输出，第二层反序输出，第三层正序输出，以此类推。做法有两种：一、仍然采用level-travel，只是引入一个标记，判断是否反转得到的数列； 二、考虑到stack的特点，利用stack FILO的特点来直接输出；两种方法都贴出来

利用stack的：

class Solution {public:    vector<vector<int> > zigzagLevelOrder(TreeNode *root) {        bool isRe = false;        vector<int> a;        stack<TreeNode *> s1, s2;                if (root == NULL)            return ret;        s1.push(root);        while (!s1.empty()){                TreeNode *tmp = s1.top();                s1.pop();                a.push_back(tmp->val);                    if (isRe){if (tmp->right)s2.push(tmp->right);if (tmp->left)s2.push(tmp->left);}else{if (tmp->left)s2.push(tmp->left);if (tmp->right)s2.push(tmp->right);}                if (s1.empty()){                    ret.push_back(a);                    isRe = !isRe;swap(s1, s2);                    a.clear();}                }        return ret;    }private:    vector<vector<int>> ret;};

利用queue的,这里由于引入了swap，所以可以复用同一个代码流程，代码会短一些;

class Solution {public:    vector<vector<int> > zigzagLevelOrder(TreeNode *root) {        vector<vector<int>> ret;        queue<TreeNode *> current, next;   //利用两个队列的交替来区分每一层        bool isRe = false;        vector<int> v;                if (root == NULL)            return ret;        current.push(root);        while (!current.empty()){            TreeNode *tmp = current.front();            current.pop();            v.push_back(tmp->val);                        if (tmp->left)                next.push(tmp->left);            if (tmp->right)                next.push(tmp->right);                            if(current.empty()){                if (isRe){                    reverse(v.begin(), v.end());                }                ret.push_back(v);                swap(current,next);                isRe = !isRe;                v.clear();            }        }            }};