leetCode . Binary Tree Inorder/ Preorder/ Post Traversal iteratively

最新推荐文章于 2026-01-03 03:21:30 发布

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本文详细介绍了二叉树的中序、前序及后序遍历算法，并提供了基于Morris遍历的具体实现，该方法能有效降低空间复杂度至O(1)，特别适合于需要高效遍历二叉树的场景。

问题

https://leetcode.com/problems/binary-tree-inorder-traversal/
https://leetcode.com/problems/binary-tree-postorder-traversal/
https://leetcode.com/problems/binary-tree-preorder-traversal/

解法

http://www.cnblogs.com/AnnieKim/archive/2013/06/15/morristraversal.html

Inorder 时间复杂度 3N 空间复杂度 O（1）

/**
 * 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> inorderTraversal(TreeNode* root) {
        vector<int> ret;
        TreeNode* cur = root;
        while(cur != NULL)
        {
            if (cur->left == NULL)
            {
                ret.push_back(cur->val);
                cur = cur->right;
            }
            else
            {
                TreeNode * pre = cur->left;
                while(pre->right != NULL && pre->right != cur)
                    pre = pre->right;
                if (pre->right == NULL)
                {
                    pre->right = cur;
                    cur = cur->left;
                }
                else
                {
                    pre->right = NULL;
                    ret.push_back(cur->val);
                    cur = cur->right;
                }
            }

        }
        return ret;
    }
};

Preorder 时间复杂度 3N 空间复杂度 O（1）

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