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  • 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}".
     
     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].

    和上面要求一样,只是要返回以中序方式序列的元素,这次用递归实现:

     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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  • 原文地址:https://www.cnblogs.com/fanchangfa/p/4041649.html
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