Binary Tree Inorder/Preorder Traversal 返回中序和前序/遍歷二叉樹的元素集合

範長法@三月軟體發表於2014-10-21
二叉樹

給定一個二叉樹，以集合方式返回其中序/先序方式遍歷的所有元素。

有兩種方法，一種是經典的中序/先序方式的經典遞迴方式，另一種可以結合棧來實現非遞迴

 

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}".
 
 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     vector<int> inorderTraversal(TreeNode *root) {
13         vector<int> ret;
14         if(root == NULL)
15             return ret;
16             
17         stack<TreeNode *> stack;
18         stack.push(root);
19         
20         while(!stack.empty()){
21             TreeNode *node = stack.top();
22             stack.pop();
23             if(node->left == NULL && node->right == NULL){
24                 ret.push_back(node->val);
25             }
26             else{
27                 if(node->right != NULL)
28                     stack.push(node->right);
29                 stack.push(node);
30                 if(node->left != NULL)
31                     stack.push(node->left);
32                     
33                 node->left = node->right = NULL;
34             }
35             
36         }
37         
38         return ret;
39             
40     }
41 };

 

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].

和上面要求一樣，只是要返回以中序方式序列的元素，這次用遞迴實現：

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     vector<int> preorderTraversal(TreeNode *root) {
13         vector<int> ret;
14         if(root == NULL)
15             return ret;
16         PreorderTraversal(root,ret);
17         return ret;
18     }
19     
20     void PreorderTraversal(TreeNode *root,vector<int> &ret){
21         if(root != NULL){
22             ret.push_back(root->val);
23             PreorderTraversal(root->left,ret);
24             PreorderTraversal(root->right,ret);
25         }
26     }
27 };

 

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