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InorderTraversal.cpp
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InorderTraversal.cpp
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/**
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].
Note: Recursive solution is trivial, could you do it iteratively?
confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
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 a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
#include <stdlib.h>
#include <iostream>
#include <vector>
using std::cout;
using std::endl;
using std::cin;
using std::vector;
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) {
std::vector<int> result;
if(root == NULL){
} else{
addNode(result,root);
}
return result;
}
void addNode(vector<int> &result, TreeNode *node){
if(node != NULL){
if(node -> left != NULL){
addNode(result, node -> left);
}
result.push_back(node -> val);
if(node -> right != NULL){
addNode(result, node -> right);
}
} else{
/*nothing to do*/
}
}
};
int main(int argc, char const *argv[])
{
Solution test;
//construct the tree
TreeNode *first = new TreeNode(1);
TreeNode *second = new TreeNode(2);
TreeNode *third = new TreeNode(3);
TreeNode *forth = new TreeNode(4);
TreeNode *fifth = new TreeNode(5);
first -> left = second;
first -> right = third;
third -> left = forth;
forth -> right = fifth;
vector<int> result = test.inorderTraversal(first);
for(int i = 0; i != result.size(); i ++){
cout << result[i] << endl;
}
return 0;
}