[leetcode] 103. Binary Tree Zigzag Level Order Traversal 解题报告

题目链接：https://leetcode.com/problems/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,#,#,15,7},

    3   / \  9  20    /  \   15   7

return its zigzag level order traversal as:

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

思路：可以用两个栈来分别保存奇数层和偶数层的结点，奇数层是从左到右，偶数层是从右到左。按照栈的后入先出的特性，在将结点加入到奇数层的栈的时候应该用右向左加，即先入栈根的右结点，再入栈根的左节点，这样计数层出栈的时候就会按照从左到右的顺序出栈。保存偶数层栈的顺序正好相反。

代码如下：

/** * 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) {        if(!root) return result;        st1.push(root);        while(!st1.empty())        {            vector<int> tem;            while(!st1.empty())//奇数层            {                TreeNode* node = st1.top();                tem.push_back(node->val);                st1.pop();                if(node->left) st2.push(node->left);                if(node->right) st2.push(node->right);            }            result.push_back(tem);            tem.clear();            while(!st2.empty())//偶数层            {                TreeNode* node = st2.top();                tem.push_back(node->val);                st2.pop();                if(node->right) st1.push(node->right);                if(node->left) st1.push(node->left);            }            if(tem.size() != 0)                 result.push_back(tem);        }        return result;    }private:    vector<vector<int>> result;    stack<TreeNode*> st1;    stack<TreeNode*> st2;};

还有一种时间复杂度O(n^2)的解法, 就是队列层次遍历, 然后看当前是奇数层还是偶数层, 如果是奇数层就正向加到数组中, 如果是偶数层就加到数组头部. 简单粗暴.

代码如下:

/** * 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) {        if(!root) return {};        queue<pair<TreeNode*, int>> que;        que.push(make_pair(root, 1));        vector<vector<int>> result;        vector<int> vec;        while(!que.empty())        {            auto val = que.front();            que.pop();            if(val.first->left) que.push(make_pair(val.first->left, val.second+1));            if(val.first->right) que.push(make_pair(val.first->right, val.second+1));            if(val.second%2 == 1) vec.push_back(val.first->val);            else vec.insert(vec.begin(), val.first->val);            if(que.empty() || val.second != que.front().second)            {                result.push_back(vec);                vec.clear();            }        }        return result;    }};