[leetcode] 103. 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],

    3   / \  9  20    /  \   15   7

return its zigzag level order traversal as:

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

解法一：

加一个flag变量记录是否reverse。

/** * 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;        if(!root) return res;        queue<TreeNode*> q;        q.push(root);                bool reverse_flag = false;        while(!q.empty()){            int sz = q.size();            vector<int> level;            for(int i=0; i<sz; i++){                TreeNode* tmp = q.front();                q.pop();                level.push_back(tmp->val);                if(tmp->left) q.push(tmp->left);                if(tmp->right) q.push(tmp->right);            }            if(reverse_flag){                reverse(level.begin(),level.end());            }            res.push_back(level);            reverse_flag = !reverse_flag;        }        return res;            }};

解法二：

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