Binary Tree Inorder Traversal [leetcode] 非递归的三种解法

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第一种方法是Morris Traversal

是O(n)时间复杂度,且不需要额外空间的方法。缺点是需要修改树。

通过将叶子节点的right指向其中序后继。

代码如下

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

第二种方法利用栈,模拟递归过程

    vector<int> inorderTraversal(TreeNode *root) {        vector<int> res;        vector<TreeNode*> stack;        stack.push_back(root);        set<TreeNode*> visited;        while (stack.size())        {            TreeNode * cur = stack.back();            stack.pop_back();            if (!cur) continue;            if (visited.find(cur) == visited.end())//visited for the first time            {                stack.push_back(cur->right);                stack.push_back(cur);                stack.push_back(cur->left);                visited.insert(cur);            }            else//visited for the second time                res.push_back(cur->val);        }        return res;    }


第三种方法来自《数据结构》

将所有左子节点入栈,当没有左子结点时取栈顶元素打印并转移到右子节点

    vector<int> inorderTraversal(TreeNode *root) {        vector<int> res;        vector<TreeNode*> stack;        TreeNode* cur = root;        while (stack.size() || cur)        {            if (cur)            {                stack.push_back(cur);                cur = cur->left;            }            else            {                cur = stack.back();                stack.pop_back();                res.push_back(cur->val);                cur = cur->right;            }        }        return res;    }


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