Binary Tree Inorder/Preorder Traversal 返回中序和前序/遍历二叉树的元素集合

给定一个二叉树，以集合方式返回其中序/先序方式遍历的所有元素。

有两种方法，一种是经典的中序/先序方式的经典递归方式，另一种可以结合栈来实现非递归

Given a binary tree, return the inorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},

   1
    \
     2
    /
   3

return [1,3,2].

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1
  / \
 2   3
    /
   4
    \
     5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

 

 /**
  * 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> inorderTraversal(TreeNode *root) {
         vector<int> ret;
         if(root == NULL)
             return ret;

         stack<TreeNode *> stack;
         stack.push(root);

         while(!stack.empty()){
             TreeNode *node = stack.top();
             stack.pop();
             if(node->left == NULL && node->right == NULL){
                 ret.push_back(node->val);
             }
             else{
                 if(node->right != NULL)
                     stack.push(node->right);
                 stack.push(node);
                 if(node->left != NULL)
                     stack.push(node->left);

                 node->left = node->right = NULL;
             }

         }

         return ret;

     }
 };

Given a binary tree, return the preorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},

   1
    \
     2
    /
   3

return [1,2,3].

和上面要求一样，只是要返回以中序方式序列的元素，这次用递归实现：

 /**
  * 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) {
         vector<int> ret;
         if(root == NULL)
             return ret;
         PreorderTraversal(root,ret);
         return ret;
     }

     void PreorderTraversal(TreeNode *root,vector<int> &ret){
         if(root != NULL){
             ret.push_back(root->val);
             PreorderTraversal(root->left,ret);
             PreorderTraversal(root->right,ret);
         }
     }
 };