题解 | #二叉树的后序遍历#非递归，单栈

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param root TreeNode类 
     * @return int整型vector
     */
     inline TreeNode* lastChild(TreeNode* root){
        if(root->right)return root->right;
        return root->left;
     }
    vector<int> postorderTraversal(TreeNode* root) {
        // write code here
        vector<int> result;
        if(!root)return result;
        deque<TreeNode*> temp={root};
        TreeNode *cur,*lastVisited,*tempNode;
        while(temp.size()){
            cur=temp.back();
            tempNode=lastChild(cur);

            if(lastVisited==tempNode||!tempNode){
                
              //  cout<<cur->val<<endl;
                result.push_back(cur->val);
                temp.pop_back();
            }else{
                if(cur->right)temp.push_back(cur->right);
                if(cur->left)temp.push_back(cur->left);
            }
            lastVisited=cur;
        }
        return result;

    }
};