Binary Tree Preorder Traversal

最新推荐文章于 2025-06-28 23:32:06 发布

原创最新推荐文章于 2025-06-28 23:32:06 发布 · 373 阅读

0 ·

CC 4.0 BY-SA版权

leetcode 同时被 2 个专栏收录

66 篇文章

订阅专栏

Algorithm

33 篇文章

订阅专栏

本文详细介绍了如何通过迭代而非递归的方式实现二叉树的前序遍历，包括Morris遍历和栈式非递归遍历两种方法，旨在提供一种在不使用额外空间的情况下进行遍历的高效算法。

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?

1. Morris前序遍历
2. 栈式非递归遍历

-------------------------

/**
 * Definition for binary tree
 * 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> ret;
        // 1
        /*
        TreeNode *cur = root;
        while (cur != NULL) {
            if (cur->left == NULL) {
                ret.push_back(cur->val);
                cur = cur->right;
            } else {
                TreeNode *l = cur->left;
                while (l->right != NULL && l->right != cur) l = l->right;
                if (l->right == NULL) {
                    l->right = cur;
                    ret.push_back(cur->val);
                    cur = cur->left;
                } else {
                    l->right = NULL;
                    cur = cur->right;
                }
            }
        }
        */
        
        // 2
        TreeNode *cur = root;
        stack<TreeNode*> tsk;
        while (cur != NULL || !tsk.empty()) {
            if (cur != NULL) {
                ret.push_back(cur->val);
                tsk.push(cur);
                cur = cur->left;
                continue;
            }
            
            cur = tsk.top()->right;
            tsk.pop();
        }
        
        return ret;
    }
};