剑指 Offer 32 - III. 从上到下打印二叉树 III（中等）

题目：

解法一：

class Solution {
public:
    void traversal(TreeNode* cur, vector<vector<int>>& result, int depth){
        if(cur==nullptr) return;
        if(result.size()==depth) result.push_back(vector<int>());
        result[depth].push_back(cur->val);
        traversal(cur->left, result, depth+1);
        traversal(cur->right, result, depth+1);
    }
    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> result;
        int depth=0;
        traversal(root, result, depth);
        for(int i=0;i<result.size();i++){                                 //取巧的方法，即将按照正常层序遍历得到的结果按层进行重排序
            if(i%2==1) reverse(result[i].begin(), result[i].end());
        }
        return result;
    }
};

解法二：
class Solution {
public:
vector<vector> zigzagLevelOrder(TreeNode* root) {
vector<vector> result;
if (!root) {
return result;
}

queue<TreeNode*> nodeQueue;
nodeQueue.push(root);
bool isOrderLeft = true; //判断插入顺序的标志位，true为从左到右，false为从右到左

while (!nodeQueue.empty()) {
deque levelList; //使用队列法的关键就在于，存储每一层元素用的是队列而不是向量，这样就可以按层调整插入的顺序
int size = nodeQueue.size();
for (int i = 0; i < size; ++i) {
auto node = nodeQueue.front(); //读取节点的顺序依然是从左到右
nodeQueue.pop();
if (isOrderLeft) {
levelList.push_back(node->val);
} else {
levelList.push_front(node->val);
}
if (node->left) {
nodeQueue.push(node->left);
}
if (node->right) {
nodeQueue.push(node->right);
}
}
result.push_back(vector{levelList.begin(), levelList.end()});
isOrderLeft = !isOrderLeft;
}

return ans;
}
};
解法二来自力扣官方题解