LeetCode: 145. Binary Tree Postorder Traversal

LeetCode: 145. Binary Tree Postorder Traversal

题目描述

Given a binary tree, return the postorder traversal of its nodes’ values.

Example:

Input: [1,null,2,3]
   1
    \
     2
    /
   3

Output: [3,2,1]

Follow up: Recursive solution is trivial, could you do it iteratively?

解题思路 —— 用STL的栈模拟系统栈

分别用两个 stack 来模拟递归计算的系统栈。

AC 代码

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution 
{
    // 递归实现
    void postorderTraversalRecursive(TreeNode* root, vector<int>& postSeq)
    {
        if(root == nullptr) return;

        postorderTraversalRecursive(root->left, postSeq);
        postorderTraversalRecursive(root->right, postSeq);
        postSeq.push_back(root->val);
    }
    // 非递归实现(模拟递归)
    void postorderTraversalWithoutRecursive(TreeNode* root, vector<int>& postSeq)
    {
        if(root == nullptr) return;

        stack<TreeNode*> stk; // 模拟递归的堆栈
        stack<int> outStk;    // 模拟输出的堆栈
        stk.push(root);

        do
        {
            // 读取参数
            TreeNode* rootParam = stk.top();
            stk.pop();

            if(rootParam == nullptr)
            {
                postSeq.push_back(outStk.top());
                outStk.pop();

                continue;
            }

            stk.push(nullptr); // 标记输出堆栈的位置

            // 反向压入输入堆栈
            if(rootParam->right != nullptr) stk.push(rootParam->right);
            if(rootParam->left != nullptr) stk.push(rootParam->left);

            //压入输出堆栈
            outStk.push(rootParam->val);
        }while(!stk.empty());
    }
public:
    vector<int> postorderTraversal(TreeNode* root)
    {
        vector<int> postSeq;
        postorderTraversalWithoutRecursive(root, postSeq);

        return postSeq;
    }
};