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No59、按之字形顺序打印二叉树

牛客网原题链接

题目描述

请实现一个函数按照之字形打印二叉树，即第一行按照从左到右的顺序打印，第二层按照从右至左的顺序打印，第三行按照从左到右的顺序打印，其他行以此类推。

示例1

输入

{8,6,10,5,7,9,11}

返回值

[[8],[10,6],[5,7,9,11]]

1、注意左右子树在两个栈中的入栈顺序

vector<vector<int> > Print(TreeNode* pRoot) {
    vector<vector<int>> result;
    if (pRoot == nullptr) return result;
    stack<TreeNode*> left_right_st;
    stack<TreeNode*> right_left_st;
    left_right_st.push(pRoot);
    while (left_right_st.size() ||  right_left_st.size()) {
        if (!left_right_st.empty()) {
            vector<int> temp;
            TreeNode* node;
            while (!left_right_st.empty()) {
                node = left_right_st.top();
                temp.push_back(node->val);
                if (node->left)//这里先左再右
                    right_left_st.push(node->left);
                if (node->right)
                    right_left_st.push(node->right);
                left_right_st.pop();
            }
            result.push_back(temp);
        }

        if (!right_left_st.empty()) {
            vector<int> temp;
            TreeNode* node;
            while (!right_left_st.empty()) {
                node = right_left_st.top();
                temp.push_back(node->val);
                if (node->right)//这里需要是先右再左
                    left_right_st.push(node->right);
                if (node->left)
                    left_right_st.push(node->left);
                right_left_st.pop();
            }
            result.push_back(temp);
        }

    }
    return result;
}

2、稍微优化一下代码

vector<vector<int> > Print(TreeNode* pRoot) {
    vector<vector<int>> result;
    if (pRoot == nullptr) return result;
    stack<TreeNode*> left_right_st;
    stack<TreeNode*> right_left_st;
    left_right_st.push(pRoot);
    while (left_right_st.size() ||  right_left_st.size()) {
        vector<int> temp;
        TreeNode* node;
        if (!left_right_st.empty()) {
            while (!left_right_st.empty()) {
                node = left_right_st.top();
                temp.push_back(node->val);
                if (node->left)
                    right_left_st.push(node->left);
                if (node->right)
                    right_left_st.push(node->right);
                left_right_st.pop();
            }
            result.push_back(temp);
            
        }
        vector<int>().swap(temp);
        if (!right_left_st.empty()) {
            while (!right_left_st.empty()) {
                node = right_left_st.top();
                temp.push_back(node->val);
                if (node->right)
                    left_right_st.push(node->right);
                if (node->left)
                    left_right_st.push(node->left);
                right_left_st.pop();
            }
            result.push_back(temp);
        }

    }
    return result;
}

3、只用一个队列来做，很不错的想法

vector<vector<int> > Print(TreeNode* pRoot) {
    vector<vector<int>> result;
    if (pRoot == nullptr) {
        return result;
    }
    queue<TreeNode*> q;
    q.push(pRoot);
    bool isLeft = false;
    while (!q.empty()) {
        int rowLen = q.size();
        vector<int> temp;
        while(rowLen--) {
            TreeNode* curNode = q.front();
            q.pop();
            if (curNode != nullptr) {
                temp.push_back(curNode->val);
                if (curNode->left)q.push(curNode->left);
                if (curNode->right)q.push(curNode->right);
            }
        }
        isLeft = !isLeft;
        if (!isLeft) {
            result.push_back(vector<int>(temp.rbegin(), temp.rend()));//注意这里是翻转一下的
        }
        else {
            result.push_back(temp);
        }
    }
    return result;
}

二刷：

1、算是二叉树的层次遍历的一种变形吧，果然还是第一反应想到这种做法

运行时间：4ms 占用内存：360k

vector<vector<int> > Print(TreeNode* pRoot) {
    if(pRoot == nullptr) return vector<vector<int>>();
    vector<vector<int>> result;
    stack<TreeNode*> left_right,right_left;
    left_right.push(pRoot);
    TreeNode*node = nullptr;
    vector<int> temp;
    while(!left_right.empty() || !right_left.empty()){
        vector<int>().swap(temp);
        while(!left_right.empty()){
            node = left_right.top();
            temp.push_back(node->val);
            left_right.pop();
            if(node->left) right_left.push(node->left);
            if(node->right) right_left.push(node->right);
        }
        if(temp.size() > 0)    result.push_back(std::move(temp));

        vector<int>().swap(temp);
        while(!right_left.empty()){
            node = right_left.top();
            temp.push_back(node->val);
            right_left.pop();
            if(node->right) left_right.push(node->right);
            if(node->left) left_right.push(node->left);

        }
        if(temp.size() > 0)   result.push_back(std::move(temp));// 可能走到头了，也就是此时temp是个空，不能把空的放在结果了
    }
    return std::move(result);
}

2、优化一下

运行时间：3ms 占用内存：504k

vector<vector<int> > Print(TreeNode* pRoot) {
    if(pRoot == nullptr) return vector<vector<int>>();
    vector<vector<int>> result;
    stack<TreeNode*> left_right,right_left;
    left_right.push(pRoot);
    TreeNode*node = nullptr;

    while(!left_right.empty() || !right_left.empty()){
        if(!left_right.empty()){
            vector<int> temp;
            while(!left_right.empty()){
                node = left_right.top();
                temp.push_back(node->val);
                left_right.pop();
                if(node->left) right_left.push(node->left);
                if(node->right) right_left.push(node->right);
            }
            result.push_back(std::move(temp));
        }

        if(!right_left.empty()){
            vector<int> temp;
            while(!right_left.empty()){
                node = right_left.top();
                temp.push_back(node->val);
                right_left.pop();
                if(node->right) left_right.push(node->right);
                if(node->left) left_right.push(node->left);

            }
            result.push_back(std::move(temp));
        }
    }
    return std::move(result);
}