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Binary Tree Level Order Traversal II

2024-09-14 23:00:47   206人阅读

方法一：最简单的方式是traversal I 中的结果直接reverse就可以

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        vector<vector<int>> result;
        if(root == nullptr)
            return result;
        stack<TreeNode *> s;
        s.push(root);
        
        while(!s.empty())
        {
            vector<TreeNode *> tmp;
            vector<int> _result;
            while(!s.empty())
            {
                tmp.push_back(s.top());
                _result.push_back(s.top()->val);
                s.pop();
            }
            
            result.push_back(_result);
            
            for(int i=tmp.size()-1; i>=0; --i)
            {
                if(tmp[i]->right != nullptr)
                    s.push(tmp[i]->right);
                if(tmp[i]->left != nullptr)
                    s.push(tmp[i]->left);
            }
        }
        
        reverse(result.begin(), result.end());
        return result;
    }
};

同样可以采用递归的方式

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        vector<vector<int>> result;
        traverse(root, 1, result);
        
        reverse(result.begin(), result.end());
        return result;
    }
    
    void traverse(TreeNode *root, int level, vector<vector<int> > &result)
    {
        if(root == nullptr)
            return;
        
        if(result.size() < level)
            result.push_back(vector<int>());
        result[level-1].push_back(root->val);
        
        traverse(root->left, level+1, result);
        traverse(root->right, level+1, result);
    }
};

方法二：这个题目除了可以使用栈数据结构，也可以使用队列数据结构

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        vector<vector<int> > result;
        if(root == nullptr)
            return result;
        
        queue<TreeNode *> que;
        que.push(root);
        
        while(!que.empty())
        {
            vector<TreeNode *> tmp;
            vector<int> _result;
            while(!que.empty())
            {
                tmp.push_back(que.front());
                _result.push_back(que.front()->val);
                que.pop();
            }
            
            result.push_back(_result);
            for(int i=0; i<tmp.size(); ++i)
            {
                if(tmp[i]->left != nullptr)
                    que.push(tmp[i]->left);
                if(tmp[i]->right != nullptr)
                    que.push(tmp[i]->right);
            }
        }
        
        reverse(result.begin(), result.end());
        return result;
    }
};

 

Binary Tree Level Order Traversal II


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