[leetcode-145-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>result;
        if (root == NULL)return result;
        stack<TreeNode*>st;
        bool flag;
        TreeNode* temp;
        do
        {
            while (root != NULL)//左结点入栈
            {
                st.push(root);
                root = root->left;
            }
            flag = true;//设置root 为已经访问过
            temp = NULL;//temp 指向栈顶结点的前一个已访问的结点

            while (!st.empty() && flag)
            {
                root = st.top();
                if (root ->right == temp)//右孩子不存在或者右孩子已经访问过 则访问 root结点
                {
                    result.push_back(root->val);//访问
                    st.pop();
                    temp = root;//指向被访问的结点
                }
                else
                {
                    root = root->right;
                    flag = false;//设置为未被访问过
                }
            }
        } while (!st.empty());
        return result;
    }

 

转载于:https://www.cnblogs.com/hellowooorld/p/6442169.html