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);
        }
    }
};