[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