二叉树遍历非递归程序 -- 使用栈模拟系统栈

宇宙之一粟

发布于 2020-10-26 10:39:23

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发布于 2020-10-26 10:39:23

文章被收录于专栏：宇宙之_一粟

一、前序遍历

给定一个二叉树，返回它的前序遍历。

示例:

输入: [1,null,2,3]  
   1
    \
     2
    /
   3 

输出: [1,2,3]

进阶: 递归算法很简单，你可以通过迭代算法完成吗？

前序遍历C++代码

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

struct Command
{
    string s; // go, print
    TreeNode* node;
    Command(string s, TreeNode* node): s(s), node(node){}
};


class Solution {
public:
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> res;
        if (root == NULL)
        {
            return res;
        }
        
        stack<Command> stack;
        stack.push(Command("go", root));
        while ( !stack.empty() )
        {
            Command command = stack.top();
            stack.pop();

            if (command.s == "print")
            {
                res.push_back(command.node -> val);
            }
            else
            {
                assert( command.s == "go");
                if (command.node -> right)
                {
                    stack.push(Command("go", command.node -> right));
                }
                if (command.node -> left)
                {
                    stack.push(Command("go", command.node -> left));
                }
                stack.push(Command("print", command.node));
            } 
        }
        return res;
    }
};

二、中序遍历

给定一个二叉树，返回它的中序遍历。

示例:

输入: [1,null,2,3]
   1
    \
     2
    /
   3

输出: [1,3,2]

进阶: 递归算法很简单，你可以通过迭代算法完成吗？

中序遍历C++代码

// @lc code=start
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
struct Command
{
    string s; // go, print
    TreeNode *node;
    Command(string s, TreeNode *node) : s(s), node(node) {}
};

class Solution
{
public:
    vector<int> inorderTraversal(TreeNode *root)
    {
        vector<int> res;
        if (root == NULL)
        {
            return res;
        }

        stack<Command> stack;
        stack.push(Command("go", root));
        while (!stack.empty())
        {
            Command command = stack.top();
            stack.pop();

            if (command.s == "print")
            {
                res.push_back(command.node->val);
            }
            else
            {
                assert(command.s == "go");
                
                if (command.node->right)
                {
                    stack.push(Command("go", command.node->right));
                }
                stack.push(Command("print", command.node));
                if (command.node->left)
                {
                    stack.push(Command("go", command.node->left));
                }
            }
        }
        return res;
    }
};
// @lc code=end

三、后序遍历

给定一个二叉树，返回它的后序遍历。

示例:

输入: [1,null,2,3]  
   1
    \
     2
    /
   3 

输出: [3,2,1]

进阶: 递归算法很简单，你可以通过迭代算法完成吗？

后序遍历C++代码

// @lc code=start
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
struct Command
{
    string s; // go, print
    TreeNode *node;
    Command(string s, TreeNode *node) : s(s), node(node) {}
};

class Solution
{
public:
    vector<int> postorderTraversal(TreeNode *root)
    {
        vector<int> res;
        if (root == NULL)
        {
            return res;
        }

        stack<Command> stack;
        stack.push(Command("go", root));
        while (!stack.empty())
        {
            Command command = stack.top();
            stack.pop();

            if (command.s == "print")
            {
                res.push_back(command.node->val);
            }
            else
            {
                assert(command.s == "go");
                stack.push(Command("print", command.node));
                if (command.node->right)
                {
                    stack.push(Command("go", command.node->right));
                }
                if (command.node->left)
                {
                    stack.push(Command("go", command.node->left));
                }
            }
        }
        return res;
    }
};
// @lc code=end