59.按之字形顺序打印二叉树

1.思路：

层次遍历二叉树，其中用变量size来确定每一层中结点的个数，定义bool变量even来确定是第几层，若是偶数层，则将数组内的数进行颠倒，再存入二维数组中。

2.代码：

/*
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
            val(x), left(NULL), right(NULL) {
    }
};
*/
class Solution {
public:
    vector<vector<int> > Print(TreeNode* pRoot) {
        vector<vector<int>> res;
        if(pRoot == NULL) return res;
        queue<TreeNode*> q;
        q.push(pRoot);
        bool even = false;
        while(!q.empty()) {
            vector<int> tmpVector;
            const int size = q.size();  //获取当前层结点的个数
            for(int i = 0; i < size; ++i) {
                TreeNode* tmpNode = q.front();
                q.pop();
                tmpVector.push_back(tmpNode->val);
                if(tmpNode->left != NULL) {
                    q.push(tmpNode->left);
                }
                if(tmpNode->right != NULL) {
                    q.push(tmpNode->right);
                }
            }
            if(even) {
                int n = tmpVector.size();
                vector<int> tmpVector2(n);  //开辟新内存，颠倒数组
                for(int i = 0; i < n; ++i) {
                    tmpVector2[i] = tmpVector[n - 1 - i];
                }
                res.push_back(tmpVector2);
            } else {
                res.push_back(tmpVector);
            }
            even = !even;
        }
        return res;
    }
    
};

二刷代码，改进了空间复杂度，采用自定义的ReverseArr函数，对原数组进行颠倒，代码如下：

/*
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
            val(x), left(NULL), right(NULL) {
    }
};
*/
class Solution {
public:
    void ReverseArr(vector<int> &arr) {
        int tmp;
        int num = arr.size();
        for(int i = 0; i < num/2; ++i) {
            tmp = arr[i];
            arr[i] = arr[num-i-1];
            arr[num-i-1] = tmp;
        }
    }
    vector<vector<int> > Print(TreeNode* pRoot) {
        vector<vector<int>> res;
        if(pRoot == NULL) return res;
        bool l2r = true;
        queue<TreeNode*> q;
        q.push(pRoot);
        while(!q.empty()) {
            vector<int> tmpRaw;
            TreeNode *tmpNode;
            int total = q.size();
            for(int i = 0; i < total; ++i) {
                tmpNode = q.front();
                q.pop();
                if(tmpNode->left != NULL) q.push(tmpNode->left);
                if(tmpNode->right != NULL) q.push(tmpNode->right);
                tmpRaw.push_back(tmpNode->val);
            }
            if(l2r) {
                res.push_back(tmpRaw);
            } else {
                ReverseArr(tmpRaw);
                res.push_back(tmpRaw);
            }
            l2r = !l2r;
        }
        return res;
    }
};

二刷思路二：

采用两个栈来解决，其中s[0]栈用来存放当前需要打印的结点，s[1]栈用来保存下一次需要打印的结点，开始时s[0]存入根节点，s[1]入栈按照从左孩子结点到右孩子结点，这样出栈再打印就是从右往左；在s[0]为空时，将数组存入res中，并令s[0]与s[1]进行交换，s[1]出栈打印，并按照从右孩子到左孩子的顺序，入栈s[0]，这样出栈时就又变成了从左往右打印结点。如此往复，直到s[0]与s[1]同时为空，此时返回res即为最终结果。该方法利用栈的特性，省去了方法一中倒转数组的步骤，更佳！

代码二：

/*
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
            val(x), left(NULL), right(NULL) {
    }
};
*/
class Solution {
public:
    vector<vector<int> > Print(TreeNode* pRoot) {
        vector<vector<int>> res;
        if(pRoot == NULL) return res;
        stack<TreeNode*> s[2];
        int cur = 0;
        int next = 1;
        s[cur].push(pRoot);
        vector<int> tmp;
        while(!s[0].empty() || !s[1].empty()) {
            TreeNode* tmpNode = s[cur].top();
            s[cur].pop();
            tmp.push_back(tmpNode->val);
            if(cur == 0) {
                if(tmpNode->left != NULL) s[next].push(tmpNode->left);  //从左往右入栈
                if(tmpNode->right != NULL) s[next].push(tmpNode->right);
            } else {
                if(tmpNode->right != NULL) s[next].push(tmpNode->right);  //从右往左入栈
                if(tmpNode->left != NULL) s[next].push(tmpNode->left);
            }
            if(s[cur].empty()) {  //单行打印完，将tmp传入res中，并更新tmp数组和cur、next的数值
                res.push_back(tmp);
                tmp.clear();
                cur = 1 - cur;
                next = 1 - next;
            }
        }
        return res;
    }
    
};