[LeetCode] Binary Tree Zigzag Level Order Traversal（bfs）

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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]]

/** * 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> > res;        if(root==NULL)            return res;        queue<TreeNode *> q;        q.push(root);        q.push(NULL);        res = bfs(q);        return res;    }private:    vector<vector<int> > bfs(queue<TreeNode *> q){        vector<vector<int> > res;        vector<int> temp;        bool needReverse = false;                while(!q.empty()){            TreeNode *p = q.front();            q.pop();            if(p!=NULL){                temp.push_back(p->val);                if(p->left!=NULL){                    q.push(p->left);                }                if(p->right!=NULL){                    q.push(p->right);                }            }else if(p==NULL ){                               if(needReverse){                    int len = temp.size();                    for(int i=0,j=len-1;i<j;i++,j--)                        swap(temp[i],temp[j]);                }                res.push_back(temp);                temp.clear();                needReverse = (!needReverse);                if(!q.empty())                   q.push(NULL);            }                }//end while        return res;    }//end bfs};

方法：用queue實現bfs，對樹按層進行搜尋，樹的結點按層存入queue中，用NULL值標記本層的結束。用bool值按層取反決定本層的vector是否要逆置。