【LeetCode】binary-tree-level-order-traversal i&ii&zigzag

题干

binary-tree-level-order-traversal

Given a binary tree, return the level order traversal of its nodes’ values. (ie, from left to right, level by level).

For example:
Given binary tree{3,9,20,#,#,15,7},

    3   / \  9  20    /  \   15   7

return its level order traversal as:

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

binary-tree-level-order-traversal ii

Given a binary tree, return the bottom-up level order traversal of its nodes’ values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree{3,9,20,#,#,15,7},

    3   / \  9  20    /  \   15   7

return its bottom-up level order traversal as:

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

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]]

问题一：层序遍历。

问题二：层序遍历，但是要从下往上输出。

问题三：层序遍历，但是是zigzag形输出。

数据结构

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNodhttps://www.nowcoder.com/practice/d8566e765c8142b78438c133822b5118?tpId=46&tqId=29071&tPage=3&rp=3&ru=/ta/leetcode&qru=/ta/leetcode/question-ranking#e *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */

解题思路

问题一

层序遍历，利用队列维护，每次把当前队列中的结点子结点送入队列，然后弹出当前结点，每一组为一层。

问题二

根据问题一只需加入reverse反转即可。

问题三

zigzag形状输出，则只需要加一个标志位来标示本层是从后向前遍历还是从前向后，从后向前reverse就可以。

参考代码

问题一：

class Solution {public:    vector<vector<int> > levelOrder(TreeNode *root) {        vector<vector<int>>res;        if (root==NULL)            return res;        queue<TreeNode *>data;//维护结点队列        data.push(root);        while(!data.empty())        {            int size=data.size();            vector<int>cur;//存储当前层            for(int i=1;i<=size;i++)            {                TreeNode *curnode=data.front();                data.pop();                cur.push_back(curnode->val);                if (curnode->left!=NULL) data.push(curnode->left);//存储当前结点的左右孩子结点                if (curnode->right!=NULL) data.push(curnode->right);            }            res.push_back(cur);        }        return res;    }};

问题二：

class Solution {public:    vector<vector<int> > levelOrderBottom(TreeNode *root) {        vector<vector<int>>res;        if (root==NULL)            return res;        queue<TreeNode *>data;        data.push(root);        while(!data.empty())        {            int size=data.size();            vector<int>cur;            for(int i=1;i<=size;i++)            {                TreeNode *curnode=data.front();                data.pop();                cur.push_back(curnode->val);                if (curnode->left!=NULL) data.push(curnode->left);                if (curnode->right!=NULL) data.push(curnode->right);            }            res.push_back(cur);        }        reverse(res.begin(), res.end());//反转上层和底层序列        return res;    }};

问题三：

class Solution {public:    vector<vector<int> > zigzagLevelOrder(TreeNode *root) {        vector<vector<int>>res;        if (root==NULL)            return res;        queue<TreeNode *>data;        data.push(root);        int flag=0;//标志位        while(!data.empty())        {            int size=data.size();            vector<int>cur;            for(int i=1;i<=size;i++)            {                TreeNode *curnode=data.front();                data.pop();                cur.push_back(curnode->val);                if (curnode->left!=NULL) data.push(curnode->left);                if (curnode->right!=NULL) data.push(curnode->right);            }            if (flag%2==1)//根据标志位来判断是否要反转来实现zigzag                reverse(cur.begin(),cur.end());            flag++;            res.push_back(cur);        }        return res;    }};

方法讨论

三个题类似，都是树的层序遍历，主要是用队列维护，然后记录每层的结点数量进行循环就好。

易错点

reverse的使用。