二叉树

构建二叉树:

增加节点:

删除节点:

查询节点:

1.先序遍历

/* 递归版前序遍历 */
void Tree::preOrderWithRecursive(ListNode* root) {
    if (root == nullptr)return;
    std::cout << root->val << " ";
    preOrderNoRecursive(root->left);
    preOrderNoRecursive(root->right);
}

/* 非递归版前序遍历 */
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> res;
        stack<TreeNode*> s;
        s.push(root);
        while(!s.empty())
        {
            TreeNode* p = s.top();
            s.pop();
            if (p == nullptr)continue;
            res.push_back(p->val);
            s.push(p->right);
            s.push(p->left);
        }
        return res;
    }

 

2. 后序遍历

前序为 root->left->right, 后序为left->right->root

可以将前序改为root->right->left,然后进行一次翻转就是left->right->root

/* 递归版后序遍历 */
void Tree::postOrderWithRecursive(ListNode* root) {
    if (root == nullptr)return;
    postOrderNoRecursive(root->left);
    postOrderNoRecursive(root->right);
    std::cout << root->val << " ";
}
/* 非递归版后序遍历 */
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> res;
        stack<TreeNode*> s;
        s.push(root);
        while(!s.empty())
        {
            root = s.top();
            s.pop();
            if (root == nullptr)continue;
            res.push_back(root->val);
            s.push(root->left);
            s.push(root->right);
        }
        reverse(res.begin(), res.end());
        return res;
    }

 

3. 中序遍历

/* 递归版中序遍历 */
void Tree::inOrderWithRecursive(ListNode* root) {
    if (root == nullptr)return;
    inOrderNoRecursive(root->left);
    std::cout << root->val << " ";
    inOrderNoRecursive(root->right);
}

/* 非递归版中序遍历 */
vector<int> inorderTraversal(TreeNode* root) {
        vector<int> res;
        stack<TreeNode*> s;
        TreeNode* p = root;
        while(p || !s.empty())
        {
            while(p)
            {
                s.push(p);
                p = p->left;
            }
            p = s.top();
            s.pop();
            res.push_back(p->val);
            p = p->right;
        }
        return res;
    }

 

posted @ 2020-02-12 22:41  强威  阅读(129)  评论(0编辑  收藏  举报