145. Binary Tree Postorder Traversal

Posted 2021-03-21 manual-linux

• 后序遍历二叉树(非递归实现)
  题目来源

• C++代码实现

class Solution
{
public:
    vector<int> postorderTraversal(TreeNode* root)
    {
        vector<int> result;
        stack<TreeNode*> datastack1;
        stack<TreeNode*> datastack2;

        if (root != NULL)
        {
            datastack1.push(root);

            while (!datastack1.empty())
            {
                root = datastack1.top();
                datastack2.push(root);
                datastack1.pop();
                if (root->left != NULL)
                {
                    datastack1.push(root->left);
                }
                if (root->right != NULL)
                {
                    datastack1.push(root->right);
                }
            }
        }
        int length = datastack2.size();
        for (int i = 0; i < length; i++)
        {
            result.push_back((datastack2.top())->val);
            datastack2.pop();
        }

        return result;
    }
};

将最后的datastack2向result中倒数据改为如下代码也可：

        while(!datastack2.empty())
        {
            result.push_back( (datastack2.top() )->val);
            datastack2.pop();
        }
        

• 第一次出现BUG:

/**
 * 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<int> postorderTraversal(TreeNode* root)
    {
        vector<int> result;
        stack<TreeNode*> datastack1;
        stack<TreeNode*> datastack2;
        
        if(root != NULL)
        {
            datastack1.push(root);
            
            while(!datastack1.empty())
            {
                root = datastack1.top();
                datastack2.push(root);
                datastack1.pop();
                if(root->left != NULL)
                {
                    datastack1.push(root->left);
                }
                if(root->right != NULL)
                {
                    datastack1.push(root->right);
                }
            }
        }
        
        for(int i = 0;i < datastack2.size();i++)
        {
            result.push_back( (datastack2.top() )->val);
            datastack2.pop();
        }
        
        return result;
    }
};

最后发现BUG出现在

        for(int i = 0;i < datastack2.size();i++)
        {
            result.push_back( (datastack2.top() )->val);
            datastack2.pop();
        }

datastack2每pop出一次数据，datastack2.size()就会减一，同时i++，导致过早的结束了这个循环，正确的写法如下:

        int length = datastack2.size();
        for (int i = 0; i < length; i++)
        {
            result.push_back((datastack2.top())->val);
            datastack2.pop();
        }