LeetCode:Binary Tree Preorder Traversal

题目：非递归实现二叉树的前序遍历。题目链接

算法1：使用栈的非递归遍历。先用根节点初始化栈，然后循环如下操作：访问栈顶节点，先后把栈顶节点右节点和左节点压栈（次序不能反，先右节点，后左节点），代码如下：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> preorderTraversal(TreeNode *root) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        vector<int>res;
        if(root == NULL)return res;
        stack<TreeNode *> nstack;
        nstack.push(root);
        while(nstack.empty() == false)
        {
            TreeNode *p = nstack.top();
            res.push_back(p->val);
            nstack.pop();
            if(p->right)nstack.push(p->right);
            if(p->left)nstack.push(p->left);
        }
        return res;
    }
};

算法2：不使用栈的非递归前序遍历（Morris Traversal算法），只要在Morris Traversal中序遍历的算法基础上修改代码节点访问顺序即可，步骤如下，代码中红色部分是修改的

重复以下1、2直到当前节点为空。

1. 如果当前节点的左孩子为空，则输出当前节点并将其右孩子作为当前节点。

2. 如果当前节点的左孩子不为空，在当前节点的左子树中找到当前节点在中序遍历下的前驱节点（即当前节点的左子树的最右节点）。

   a) 输出当前节点。（相对中序遍历，输出位置改变了）。如果前驱节点的右孩子为空，将它的右孩子设置为当前节点（利用这个空的右孩子指向它的后缀）。当前节点更新为当前节点的左孩子。

   b) 如果前驱节点的右孩子为当前节点，将它的右孩子重新设为空（恢复树的形状）。当前节点更新为当前节点的右孩子。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> preorderTraversal(TreeNode *root) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
          TreeNode *current = root, *pre = NULL;
          vector<int> res;
          while(current != NULL)
          {                 
                if(current->left == NULL)
                {
                      res.push_back(current->val);
                      current = current->right;      
                }    
                else
                {
                      /* Find the inorder predecessor of current */
                      pre = current->left;
                      while(pre->right != NULL && pre->right != current)
                        pre = pre->right;
                        
                      if(pre->right == NULL)
                      {     /* Make current as right child of its inorder predecessor */
                            pre->right = current;
                            res.push_back(current->val);
                            current = current->left;
                      }
                      else 
                      {
                            /* Revert the changes made in if part to restore the original 
                            tree i.e., fix the right child of predecssor */   
                            pre->right = NULL;//中序是在这里输出
                            current = current->right;      
                      } 
                }
          } 
          return res;
    }
};

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