[leetcode 144]Binary Tree Preorder Traversal

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本文介绍二叉树前序遍历的三种方法：递归、栈迭代及Morris遍历。递归方法简洁但空间复杂度较高；栈迭代避免了递归的调用开销；Morris遍历则在空间复杂度上达到了最优。

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?

递归空间复杂度为O(n),时间复杂度为O(n)

/**
 * 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> res;
    vector<int> preorderTraversal(TreeNode *root) {
        if (!root) {
            return res;
        }
        res.push_back(root->val);
        preorderTraversal(root->left);
        preorderTraversal(root->right);
        return res;
    }
};

栈空间复杂度为O(n),时间复杂度为O(n)

/**
 * 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> res;
        if (!root) {
            return res;
        }
        stack<TreeNode *> p;
        p.push(root);
        while (!p.empty()) {
            auto tmp = p.top();
            p.pop();
            res.push_back(tmp->val);
            if (tmp->right) {
                p.push(tmp->right);
            }
            if (tmp->left) {
                p.push(tmp->left);
            }
        }
        return res;
    }
};

Morris二叉树遍历（参考http://www.it165.net/pro/html/201403/10857.html）空间复杂度为O(1),时间复杂度为O(n)

/**
 * 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> res;
        TreeNode *cur = root;
        while (cur) {
            if (cur->left == NULL) {
                res.push_back(cur->val);
                cur = cur->right;
            } else {
                TreeNode *node = cur->left;
                while (node->right && node->right != cur) {
                    node = node->right;
                }
                if (node->right == NULL) {
                    res.push_back(cur->val);
                    node->right = cur;
                    cur = cur->left;
                } else {
                    node->right = NULL;
                    cur = cur->right;
                }
            }
        }
        return res;
    }
};