leetcode binary-tree-zigzag-level-order-traversal

最新推荐文章于 2020-03-25 17:26:06 发布

pynash123

最新推荐文章于 2020-03-25 17:26:06 发布

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分类专栏： leetcode 算法

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本文介绍了一种特殊的二叉树遍历方法——锯齿形层级遍历，并提供了一个使用双栈实现的具体算法示例。该算法首先采用正常层序遍历方式，之后根据层级奇偶性决定是否翻转节点值顺序。

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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) {}
 * };
 */
 
 //正常层序遍历  然后每隔一层翻转
 
 
 
//两个stack
class Solution {
public:
    vector<vector<int> > zigzagLevelOrder(TreeNode *root) 
    {
        vector<vector<int> >res;
        if (NULL == root)
            return res;
        stack<TreeNode *> sltor,srtol;
        sltor.push(root);
        while(!(sltor.empty() && srtol.empty()))
        {
            vector<int> vectmp;
            if(!sltor.empty())
            {
                while(!sltor.empty())
                {
                    TreeNode * tmp = sltor.top();
                    sltor.pop();
                    vectmp.push_back(tmp->val);
                    if (NULL != tmp->left)
                        srtol.push(tmp->left);
                    if (NULL != tmp->right)
                        srtol.push(tmp->right);
                }
            }
            else
            {
                while(!srtol.empty())
                {
                    TreeNode * tmp = srtol.top();
                    srtol.pop();
                    vectmp.push_back(tmp->val);
                    if (NULL != tmp->right)
                        sltor.push(tmp->right);
                    if (NULL != tmp->left)
                        sltor.push(tmp->left);
                }
            }
            res.push_back(vectmp);
        }
        return res;
    }
};

测试

#include<iostream>
#include<vector>
#include<stack>
using namespace std;


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 (NULL == root)
            return res;
        stack<TreeNode *> sltor,srtol;
        sltor.push(root);
        while(!(sltor.empty() && srtol.empty()))
        {
            vector<int> vectmp;
            if(!sltor.empty())
            {
                while(!sltor.empty())
                {
                    TreeNode * tmp = sltor.top();
                    sltor.pop();
                    vectmp.push_back(tmp->val);
                    if (NULL != tmp->left)
                        srtol.push(tmp->left);
                    if (NULL != tmp->right)
                        srtol.push(tmp->right);
                }
            }
            else
            {
                while(!srtol.empty())
                {
                    TreeNode * tmp = srtol.top();
                    srtol.pop();
                    vectmp.push_back(tmp->val);
                    if (NULL != tmp->right)
                        sltor.push(tmp->right);
                    if (NULL != tmp->left)
                        sltor.push(tmp->left);
                }
            }
            res.push_back(vectmp);
        }
        return res;
    }
};


int main()
{
    TreeNode * root = new TreeNode(13);
    root->left = new TreeNode(9);
    root->right = new TreeNode(20);
    root->right->left = new TreeNode(15);
    root->right->right = new TreeNode(17);
    //root->left->left = new TreeNode(9);
    //root->left->right = new TreeNode(12);
    Solution s;
    vector<vector<int> > vec = s.zigzagLevelOrder(root);
    
    for(int i=0;i<vec.size();i++)
    {
        for (int j=0;j<vec[i].size();j++)
        {
            cout<<vec[i][j]<<" ";
        }
        cout<<endl;
    }
}