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?

 

思路：

用一个栈来保存节点。因为需要记录是否访问了左右子节点，所以用一个pair的第一位来记录当前的访问状态。

代码：

    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.
        vector<int> result;
        result.clear();
        stack<pair<TreeNode*, int> > treeStack;
        if(root == NULL)
            return result;
        treeStack.push(pair<TreeNode*, int>(root, 0));
        while(!treeStack.empty()){
            TreeNode *tNode = treeStack.top().first;
            int state = treeStack.top().second;
            treeStack.pop();
            if(state == 0){
                treeStack.push(pair<TreeNode*, int>(tNode, 1));
                if(tNode->left){
                    treeStack.push(pair<TreeNode*, int>(tNode->left, 0));    
                }
            }
            else if(state == 1){
                treeStack.push(pair<TreeNode*, int>(tNode, 2));
                if(tNode->right){
                    treeStack.push(pair<TreeNode*, int>(tNode->right, 0));
                }
            }
            else{
                result.push_back(tNode->val);
            }
        }
        return result;
    }