【leetcode】Binary Tree Postorder Traversal (hard) ☆

二叉树的后序遍历

用标记右子树vector的方法

vector<int> postorderTraversal(TreeNode *root) {
        vector<int> ans;
        vector<TreeNode *> stack;
        vector<bool> isRight;
        stack.push_back(root);
        isRight.push_back(false);
        TreeNode * pNode = NULL;
        while(!stack.empty())
        {
            while((pNode = stack.back()) != NULL)
            {
                pNode = pNode->left;
                stack.push_back(pNode);
                isRight.push_back(false);
            }

            while(isRight.back() == true)
            {
                stack.pop_back();
                isRight.pop_back();
                ans.push_back(stack.back()->val);
            }
            stack.pop_back();
            isRight.pop_back();

            if(!stack.empty())
            {
                pNode = stack.back();
                stack.push_back(pNode->right);
                isRight.push_back(true);
            }
        }

        return ans;
    }

仅用一个栈的方法https://oj.leetcode.com/discuss/14118/post-order-traversal-using-two-satcks

    vector<int> postorderTraversal(TreeNode *root) {
        stack<TreeNode *> st;
        vector<int> vRet;
        TreeNode *p, *pre = root;

        if (!root) return vRet;
        p = root->left;
        st.push(root);
        while (!st.empty() )
        {
            while (p) { st.push(p); p = p->left; }
            p = st.top();
            while (!p->right || pre==p->right)
            {
                vRet.push_back(p->val);
                pre = p;
                st.pop();
                if (st.empty() ) break;
                p = st.top();
            }
            p = p->right;
        }
        return vRet;
    }