leetcode：二叉樹之Binary Tree Inorder Traversal

leetcode：二叉樹之Binary Tree Inorder Traversal

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?

即，二叉樹中序遍歷

c++實現：

#include <iostream>
#include <vector>
#include <malloc.h>
#include <stack>

using namespace std;
struct TreeNode 
{
	int val;
	TreeNode *left;
	TreeNode *right;
	TreeNode(int x) : val(x), left(NULL), right(NULL) { }
};
void CreateBiTree(TreeNode* &T)
{
	char ch;
	cin>>ch;

	if(ch=='#') 
		T=NULL;
	else
	{
		T = (TreeNode*)malloc(sizeof(TreeNode));
		if(!T) exit(0);
		T->val = ch-'0';
		CreateBiTree(T->left);  
		CreateBiTree(T->right); 
	}
}
//使用棧
vector<int> inorderTraversal(TreeNode *root)
{
	vector<int> result;
	const TreeNode *p = root;
	stack<const TreeNode *> s;
	while (!s.empty() || p != NULL) 
	{
		if (p != NULL) 
		{
			s.push(p);
			p = p->left;
		} else{
			p = s.top();
			s.pop();
			result.push_back(p->val);
			p = p->right;
		}
	}
	return result;
}
int main()
{
	TreeNode* root(0);
	CreateBiTree(root);

	cout<<"輸出中序遍歷結果："<<endl;
	vector<int> v;
	v=inorderTraversal(root);
	for(int i = 0;i < v.size();i++)
        cout<<v[i];
	cout<<endl;
    
	return 0;
}

測試結果：