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?

vector<int> postorderTraversal(TreeNode *root){
    vector<int> res;
    if(root == NULL) return res;
    stack<TreeNode *> record;
    record.push(root);
    TreeNode *pre = NULL;
    while(!record.empty()){
        TreeNode *node = record.top();
        if((node->left == NULL  && node->right == NULL)||(pre!=NULL &&(pre == node->left || pre == node->right)) ){
            res.push_back(node->val);
            record.pop();
            pre = node;
        }else{
            if(node->right) record.push(node->right);
            if(node->left ) record.push(node->left);
        }    
    }
    return res;
}

 另一種方法：

通過不斷的交換左右孩子，然後壓棧，最後彈出，即可得到結果

時間複雜度為O(h),h為樹的高度，空間複雜度為O(n)

vector<int> postorderTraversa(TreeNode *root){
    vector<int> res;
    if(root == NULL) return res;
    stack<TreeNode *> post_record,reverse_record;
    post_record.push(root);
    while(!post_record.empty()){
        TreeNode *node = post_record.top();
        reverse_record.push(node);
        post_record.pop();
        if(node->left) post_record.push(node->left);
        if(node->right) post_record.push(node->right);
    }
    while(!reverse_record.empty()){
        res.push_back(reverse_record.top()->val);
        reverse_record.pop();
    }
    return res;
}