题解 | #二叉树的后序遍历#

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
#include <vector>
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
     */
    vector<int> postorderTraversal(TreeNode* root) {
        // write code here
        vector<int> order;
        stack<int> rootval;
        stack<TreeNode*> Node;
        while(root!=nullptr||!Node.empty()){
            if(root!=nullptr){
                rootval.push(root->val);
                Node.push(root);
                root = root->right;
            }
            else{
                root = Node.top();
                Node.pop();
                root = root->left;
            }
        }
        while(!rootval.empty()){
            order.push_back(rootval.top());
            rootval.pop();
        }
        return order;
    }
    private:
    void postorder(vector<int> &order,TreeNode* root){
        if(root == nullptr)
            return;
        postorder(order, root->left);
        postorder(order, root->right);
        order.push_back(root->val);
    }
};

后序遍历其实可以看成中右左的逆序。就是把中序的左右遍历反一下，将结果存到栈里，再依次弹出打印即可