题解 | #实现二叉树先序，中序和后序遍历#

/**
 * struct TreeNode {
 *    int val;
 *    struct TreeNode *left;
 *    struct TreeNode *right;
 * };
 */

class Solution {
public:
    /**
     * 
     * @param root TreeNode类 the root of binary tree
     * @return int整型vector<vector<>>
     */
    vector<vector<int> > threeOrders(TreeNode* root) {
        // write code here
        vector<vector<int>> res;
        res.push_back(preorder(root));
        res.push_back(inorder(root));
        res.push_back(postorder(root));
        return res;
    }
    vector<int> preorder(TreeNode* root) {
        if (!root) return {};
        vector<int> res;
        stack<TreeNode*> s;
        s.push(root);
        while (!s.empty()) {
            TreeNode* node = s.top();
            s.pop();
            res.push_back(node->val);
            if (node->right) s.push(node->right);
            if (node->left) s.push(node->left);
        }
        return res;
    }
    vector<int> inorder(TreeNode* root) {
        if (!root) return {};
        vector<int> res;
        stack<TreeNode*> s;
        TreeNode* cur = root;
        while (!s.empty() || cur) {
            while (cur) {
                s.push(cur);
                cur = cur->left;
            } 
            cur = s.top();
            s.pop();
            res.push_back(cur->val);
            cur = cur->right;
        }
        return res;
    }
    //后序可以先按根右左的顺序便利，在翻转结果即可
    vector<int> postorder(TreeNode* root) {
        if (!root) return {};
        vector<int> res;
        stack<TreeNode*> s;
        s.push(root);
        while (!s.empty()) {
            TreeNode* node = s.top();
            s.pop();
            res.push_back(node->val);
            if (node->left) s.push(node->left);
            if (node->right) s.push(node->right);
        }
//         for (int i = 0, j = res.size() - 1; i < j; )
//             swap(res[i++], res[j--]);
        reverse(res.begin(), res.end());
        return res;
    }
};