Leetcode Binary Tree Preorder Traversal

最新推荐文章于 2025-08-04 20:35:07 发布

rosepicker

最新推荐文章于 2025-08-04 20:35:07 发布

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文章标签： leetcode

本文链接：https://blog.youkuaiyun.com/rosepicker/article/details/74941020

本文介绍了一种实现二叉树前序遍历的方法，包括递归和迭代两种方式。递归方法直观且易于理解，而迭代方法则通过使用栈来避免递归带来的开销。

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Given a binary tree, return the preorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},

return [1,2,3].

Note: Recursive solution is trivial, could you do it iteratively?

递归大法代码如下：

/**
 * 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> preorderTraversal(TreeNode* root) {
        vector<int> result;
        preorder(root,result);
        return result;
    }
    
    void preorder(TreeNode* root,vector<int>& result)
    {
        if(root == NULL)
            return;
        result.push_back(root->val);
        preorder(root->left,result);
        preorder(root->right,result);
    }
};

迭代大法代码如下：

/**
 * 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> preorderTraversal(TreeNode* root) {
        vector<int> result;
        stack<TreeNode*> data;
        while(root)
        {
            if(root->right)
                data.push(root->right);
            result.push_back(root->val);
            root = root->left;
            if(root == NULL && !data.empty())
            {
                root = data.top();
                data.pop();
            }
        }
        return result;
    }
};