leetcode 二叉树的遍历栈的实现 Preorder Traversal Inorder Traversal Postorder Traversal_given the preorder traversal and inorder traversal

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].

Note: Recursive solution is trivial, could you do it iteratively?

 class Solution {
 public:
	 vector<int> preorderTraversal(TreeNode* root) {
		 vector<int> res;
		
		 if(!root) return res;
		 stack<TreeNode*> S;
		 S.push(root);
		 TreeNode* node;
		 while (!S.empty())
		 {
			 node = S.top();
			 S.pop();
			 res.push_back(node->val);
			 if (node->right) S.push(node->right);
			 if (node->left) S.push(node->left);
		 }
		 return res;
	 }
 };

中序遍历：

 class Solution {
 public:
	 vector<int> inorderTraversal(TreeNode* root) {
		 vector<int>res;
		 if (!root) return res;
		 stack<TreeNode *>S;
		 TreeNode* node=root;
		 while (node||!S.empty())
		 {
			 while (node)//有左就一直向左
			 {
				 S.push(node);
				 node = node->left;
			 }
			 node = S.top();
			 S.pop();
			 res.push_back(node->val);
			 node = node->right;
		 }
		 return res;
	 }
 };

后序遍历：大概就是有左右子树就push，没有左子树和右子树或者已经处理过了就输出，这里已经处理过了可以用一个child节点记录上一个输出的node,若这个child是栈顶的左孩子或者右孩子就表示已经处理过了。

class Solution {
 public:
	 vector<int> postorderTraversal(TreeNode* root) {
		 vector<int>res;
		 if (!root) return res;
		 stack<TreeNode*>S;
		 S.push(root);
		 TreeNode* node = root;
		 TreeNode* child = node;
		 while (!S.empty())
		 {
			 node = S.top();
			 if ((!node->left&&!node->right) || child == node->left || child == node->right)//左右都空了或者孩子已经处理完成了
			 {
				 res.push_back(node->val);
				 S.pop();
				 child = node;
			 }
			 else{ //先把右孩子push,再把左孩子push
				 if (node->right) S.push(node->right);
				 if (node->left)  S.push(node->left);
			 }
		 }
		 return res;
	 }
 };