三种树遍历顺序的迭代版

/**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
//迭代
vector<int> preorderTraversal(TreeNode *root) {
vector <int> InorderOfPath;
stack <TreeNode *> stack1;
if(root!=NULL){
stack1.push(root);
while(!stack1.empty()){
TreeNode * p=stack1.top();
stack1.pop();
InorderOfPath.push_back(p->val);
if(p->right) stack1.push(p->right);
if(p->left) stack1.push(p->left);
}
}
return InorderOfPath;
}

vector<int> inorderTraversal(TreeNode *root) {

vector<int> InorderPath;
stack <TreeNode *> stack1;
while(!stack1.empty() || root!=NULL){
while(root!=NULL){
stack1.push(root);
root=root->left;
}
if(!stack1.empty()){
root=stack1.top();
stack1.pop();
InorderPath.push_back(root->val);
root=root->right;//若存在右子树，则压栈进行遍历。若无，则跳过前一个while循环，直接进行弹栈操作
}
}
return InorderPath;
}

vector<int> postorderTraversal(TreeNode* root){
vector<int> path;
stack<TreeNode *> stack1;
TreeNode * cur;
TreeNode * prev;
cur=root;
prev=NULL;
while(cur){
stack1.push(cur);
cur=cur->left;
}//同样是遍历压栈到最左边的结点
while(!stack1.empty()){
cur=stack1.top();
stack1.pop();
if(cur->right==NULL || cur->right==prev){
//一个根结点被访问的前提是：无右子树或右子树已经遍历过了
path.push_back(cur->val);
prev=cur;
}
else{
stack1.push(cur);//若结点有右子树，则应再把结点压回去，等右子树遍历完再遍历这个结点
//先进入右子树
cur=cur->right;
while(cur){
stack1.push(cur);
cur=cur->left;
}
}
}
return path;
}