leetcode: 144. 二叉树的前序遍历

• 使用迭代算法
• 解题思路：Morris遍历 利用结点空指针
  

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> ans;
        while (root) {
            if (nullptr == root->left) {
                ans.emplace_back(root->val);
                root = root->right;
            } else {
                //先找到前驱节点
                TreeNode* pre = root->left;
                while (pre->right && pre->right != root) { //第一次找到前驱节点
                    pre = pre->right;
                }

                if (nullptr == pre->right) {
                    pre->right = root; //第一次来，添加回溯指针
                    ans.emplace_back(root->val);
                    root = root->left;
                } else { //第二次来 已处理完左子树
                pre->right = nullptr;//还原树的结构
                root = root->right;
                }
            }
        }
        return ans;
    }
};