二叉树的遍历实现(递归和非递归)

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二叉树的遍历(前序、中序和后序)，递归和非递归的实现。前序和中序实现基本类似，后序稍微复杂一点，需要判断当前节点的右子树是否为空或者已经访问过，代码如下：

//前序遍历:递归void preorder(TreeNode *root, vector<int> &path){    if(root != NULL)    {        path.push_back(root->val);        preorder(root->left, path);        preorder(root->right, path);    }}//中序遍历:递归void inorder(TreeNode *root, vector<int> &path){    if(root != NULL)    {        inorder(root->left, path);        path.push_back(root->val);        inorder(root->right, path);    }}//后续遍历:递归void postorder(TreeNode *root, vector<int> &path){    if(root != NULL)    {        postorder(root->left, path);        postorder(root->right, path);        path.push_back(root->val);    }}//前序遍历:非递归void preorderTraversal(TreeNode *root, vector<int> &path){    stack<TreeNode *> s;    TreeNode *p = root;    while(p != NULL || !s.empty())    {        while(p != NULL)        {            path.push_back(p->val);            s.push(p);            p = p->left;        }        if(!s.empty())        {            p = s.top();            s.pop();            p = p->right;        }    }}//中序遍历:非递归void inorderTraversal(TreeNode *root, vector<int> &path){    stack<TreeNode *> s;    TreeNode *p = root;    while(p != NULL || !s.empty())    {        while(p != NULL)        {            s.push(p);            p = p->left;        }        if(!s.empty())        {            p = s.top();            path.push_back(p->val);            s.pop();            p = p->right;        }    }}//后序遍历:非递归第一种void postorderTraversal1(TreeNode *root, vector<int> &path){    stack<TempNode *> s;    TreeNode *p = root;    TempNode *temp;    while(p != NULL || !s.empty())    {        while(p != NULL) //沿左子树一直往下搜索，直至出现没有左子树的结点        {            TempNode *tempNode = new TempNode;            tempNode->btnode = p;            tempNode->isFirst = true;            s.push(tempNode);            p = p->left;        }        if(!s.empty())        {            temp = s.top();            s.pop();            if(temp->isFirst == true)   //表示是第一次出现在栈顶            {                temp->isFirst = false;                s.push(temp);                p = temp->btnode->right;            }            else  //第二次出现在栈顶            {                path.push_back(temp->btnode->val);                p = NULL;            }        }    }}//后序遍历:非递归第二种void postorderTraversal2(TreeNode *root, vector<int> &path){    stack<TreeNode *> s;    TreeNode *curr = NULL;    TreeNode *last = NULL;s.push(root);    while(!s.empty())    {curr = s.top();if((curr->lchild == NULL && curr->rchild == NULL) ||   (last != NULL && (last == curr->lchild || last == curr->rchild))) {path.push_back(curr->val);s.pop();last = curr;}else{ if(curr->rchild != NULL) s.push(curr->rchild); if(curr->lchild != NULL) s.push(curr->lchild);}    }}

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