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

題幹

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]
]

問題一：層序遍歷。

問題二：層序遍歷，但是要從下往上輸出。

問題三：層序遍歷，但是是zigzag形輸出。

數據結構

/**
 * 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) {}
 * };
 */

解題思路

問題一

層序遍歷，利用隊列維護，每次把當前隊列中的結點子結點送入隊列，然後彈出當前結點，每一組爲一層。

問題二

根據問題一隻需加入reverse反轉即可。

問題三

zigzag形狀輸出，則只需要加一個標誌位來標示本層是從後向前遍歷還是從前向後，從後向前reverse就可以。

參考代碼

問題一：

class Solution {
public:
    vector<vector<int> > levelOrder(TreeNode *root) {
        vector<vector<int>>res;
        if (root==NULL)
            return res;
        queue<TreeNode *>data;//維護結點隊列
        data.push(root);
        while(!data.empty())
        {
            int size=data.size();
            vector<int>cur;//存儲當前層
            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);//存儲當前結點的左右孩子結點
                if (curnode->right!=NULL) data.push(curnode->right);
            }
            res.push_back(cur);
        }
        return res;
    }
};

問題二：

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());//反轉上層和底層序列
        return res;
    }
};

問題三：

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;//標誌位
        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)//根據標誌位來判斷是否要反轉來實現zigzag
                reverse(cur.begin(),cur.end());
            flag++;
            res.push_back(cur);
        }
        return res;
    }
};

方法討論

三個題類似，都是樹的層序遍歷，主要是用隊列維護，然後記錄每層的結點數量進行循環就好。

易錯點

reverse的使用。