[LeetCode]Binary Tree Postorder Traversal

Given a binary tree, return the postorder traversal of its nodes' values.

For example: Given binary tree {1,#,2,3},

   1
    \
     2
    /
   3

 

return [3,2,1].

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

 

思考：原思路是为每个节点增加一个标记。这个思路先序遍历不要标记，因为当前访问结点直接出栈，不用看是否访问过。不过此思路redefinition of 'struct TreeNode'。

struct TreeNode {
	int val;
	TreeNode *left;
	TreeNode *right;
	bool flag;
	TreeNode(int x) : val(x), left(NULL), right(NULL), flag(false) {}
};
 
class Solution {
private:
	vector<int> ret;
public:
    vector<int> postorderTraversal(TreeNode *root) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
		ret.clear();
		if(!root) return ret;
		stack<TreeNode *> s;
		s.push(root);
		while(!s.empty())
		{
			TreeNode *cur=s.top();
			if(cur->left&&!cur->left->flag)
				s.push(cur->left);
			else if(cur->right&&!cur->right->flag) 
				s.push(cur->right);
			else
			{
				ret.push_back(cur->val);
				cur->flag=true;
				s.pop();
			}
		}
		return ret;
    }
};

       既然不能增加标记结点，那么遍历过的结点直接“删掉”。不过这么做破坏了原树结构。　

class Solution {
private:
	vector<int> ret;
public:
    vector<int> postorderTraversal(TreeNode *root) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
		ret.clear();
		if(!root) return ret;
		stack<TreeNode *> s;
		s.push(root);
		TreeNode *cur=s.top();
		while(!s.empty())
		{
			TreeNode *cur=s.top();
			if(!cur->left&&!cur->right)
			{
				ret.push_back(cur->val);
				s.pop();
			}
			if(cur->right)
			{
				s.push(cur->right);
				cur->right=NULL;
			}	
			if(cur->left)
			{
				s.push(cur->left);
				cur->left=NULL;
			}	
		}
		return ret;
    }
};

      网上搜索有没有更好的方法，发现自己思维定势了，只知道不能递归就用stack代替。参考http://www.cnblogs.com/changchengxiao/p/3416402.html