LeetCode题解： Binary Tree Zigzag Level Order Traversal

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：

通过迭代方法按行访问。然后每访问一行，改变一次访问顺序和存放子结点的顺序。

题解：

/** * Definition for binary tree * 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>> ret;                if (root == nullptr)            return ret;                // direction of visiting        bool fwd_direction = false;                list<TreeNode*> this_row;        list<TreeNode*> next_row;                this_row.push_back(root);        ret.push_back(vector<int>());                while(!this_row.empty())        {            TreeNode* l;                        if (fwd_direction)             {                l = this_row.front();                this_row.pop_front();            }            else            {                l = this_row.back();                this_row.pop_back();            }                        ret.back().push_back(l->val);                        if (fwd_direction)            {                if (l->right != nullptr) next_row.push_back(l->right);                if (l->left != nullptr) next_row.push_back(l->left);            }            else            {                if (l->left != nullptr) next_row.push_front(l->left);                if (l->right != nullptr) next_row.push_front(l->right);            }                        if (this_row.empty())            {                swap(this_row, next_row);                ret.push_back(vector<int>());                fwd_direction = !fwd_direction;            }        }                //if (ret.back().empty())         ret.pop_back();                return ret;    }};