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dL Zu 算法学习之运用栈模拟递归-二叉树的中序遍历 - i'm jackey - i'm jackey

算法学习之运用栈模拟递归-二叉树的中序遍历

Jackey C/C++ 30 次浏览 , , , 没有评论

题目要求:

给定一个二叉树的根节点 root ,返回它的 中序 遍历。

 

示例 1:

输入:root = [1,null,2,3]
输出:[1,3,2]
示例 2:

输入:root = []
输出:[]
示例 3:

输入:root = [1]
输出:[1]
示例 4:

输入:root = [1,2]
输出:[2,1]
示例 5:

输入:root = [1,null,2]
输出:[1,2]
提示:

树中节点数目在范围 [0, 100] 内
-100 <= Node.val <= 100
进阶: 递归算法很简单,你可以通过迭代算法完成吗?

来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/binary-tree-inorder-traversal

解题代码:

//Definition for a binary tree node.
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode() : val(0), left(nullptr), right(nullptr) {}
    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
    TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};

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) { // 中序遍历,先访问左孩子,再打印节点,再访问右孩子
        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));
                // 先推入访问右孩子,再打印
                stack.push(Command("print", command.node));
                // 实际执行,先推入左孩子,再打印,再访问右孩子
                if (command.node -> left)
                    stack.push(Command("go", command.node->left));

            }
        }
        return res;
    }
};

 

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