Binary Tree Zigzag Level Order Traversal——关于广度优先的经典面试题

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]
]

这道题明显是要用广度优先算法来实现。

需要注意的是：

深度优先算法要用栈来实现，广度优先需要用队列来实现。之前都是用深度优先算法，这是第一次写关于广度优先算法的实例。

PS:写代码时一定要注意复制粘贴所带来的错误·····················

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
struct Node{
    TreeNode* node;
    int level;
    Node(){};
    Node(TreeNode* root,int lev):node(root),level(lev){};
};
class Solution {
public:
    vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
        vector<vector<int>> res;
        vector<int> tempRes;
        if(root==NULL)
            return res;
        queue<Node> Que;
        Que.push(Node(root,));
        int curLevel=;
        while(!Que.empty())
        {
            Node front=Que.front();
            if(front.node->left!=NULL)
                Que.push(Node(front.node->left,front.level+));
            if(front.node->right!=NULL)
                Que.push(Node(front.node->right,front.level+));
            if(curLevel==front.level)
            {
                tempRes.push_back(front.node->val);
            }
            else
            {
                if(curLevel%==)
                    reverse(tempRes.begin(),tempRes.end());
                res.push_back(tempRes);
                tempRes.clear();
                curLevel=front.level;
                tempRes.push_back(front.node->val);
            }
            Que.pop();
        }
        if(curLevel%==)
                reverse(tempRes.begin(),tempRes.end());
        res.push_back(tempRes);
        tempRes.clear();
        return res;

    }
};

　　网上看到另一种解法，没有定义结构体，设置一个标志来表明是从左往右还是从右往左。这里用的是一个枚举类型enum{L,R};每遍历一层，对方向进行一个转变。

/**
 * 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>>result;
    if (!root)
        return result;
    vector<int>vec;
    queue<TreeNode*>q1;
    TreeNode *temp = root;
    enum Dir{L,R};
    Dir dir = L;
    q1.push(root);  

    while (!q1.empty())
    {
        queue<TreeNode *>q2;
        while (!q1.empty())
        {
            temp = q1.front();
            q1.pop();
            if (temp->left)
                q2.push(temp->left);
            if (temp->right)
                q2.push(temp->right);
            vec.push_back(temp->val);
        }
        if (dir == R)
        {
            reverse(vec.begin(), vec.end());
            dir = L;
        }
        else
            dir = R;
        result.push_back(vec);
        vec.clear();
        q1 = q2;
    }
    return result;
    }
};