题解 | #按之字形顺序打印二叉树#

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
#include <stack>
#include <vector>
using namespace std;
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param pRoot TreeNode类 
     * @return int整型vector<vector<>>
     */
    vector<vector<int> > Print(TreeNode* pRoot) {
        // write code here
        vector<vector<int> > res;

        if (pRoot == nullptr)
            return res;
        
        stack<TreeNode*> levels[2];
        int current = 0;
        int next = 1;

        levels[current].push(pRoot);
        while (!levels[0].empty() || !levels[1].empty()) {
            
            vector<int> levelValues;
            while (!levels[current].empty()) {
                TreeNode *pNode = levels[current].top();
                levels[current].pop();

                levelValues.push_back(pNode->val);

                if (current == 0) {
                    if (pNode->left != nullptr)
                        levels[next].push(pNode->left);
                    if (pNode->right != nullptr)
                        levels[next].push(pNode->right);
                } else {
                    if (pNode->right != nullptr)
                        levels[next].push(pNode->right);
                    if (pNode->left != nullptr)
                        levels[next].push(pNode->left);
                }
            }

            res.push_back(levelValues);
            levelValues.clear();

            if (levels[current].empty()) {
                printf("\n");
                current = 1 - current;
                next = 1 - next;
            }

        }
        return res;
    }
};