5.1.3—二叉树的遍历—Binary Tree Postorder Traversal

描述
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?

#include "BinaryTree.h"#include <stack>#include<vector>using namespace std;//===二叉树的后序遍历，递归版本void PostorderTraversal(BinaryTreeNode *proot){if (proot){if (proot->m_pLeft)PostorderTraversal(proot->m_pLeft);if (proot->m_pRight)PostorderTraversal(proot->m_pRight);cout << proot->m_nValue << " ";}}//===二叉树的后序遍历，迭代版本void PostorderTraversal2(BinaryTreeNode *proot){stack<BinaryTreeNode*> temp;vector<int> cahe;BinaryTreeNode *p = proot,*q;bool flag;do{while (p){temp.push(p);p = p->m_pLeft;}flag = true;q = NULL;while ((!temp.empty()) && flag){p = temp.top();if (p->m_pRight == q){cahe.push_back(p->m_nValue);temp.pop();q = p;}else{p = p->m_pRight;flag = false;}}} while (!temp.empty());//======for (int i = 0; i < cahe.size(); i++)cout << cahe[i] << " ";cout << endl;}// ====================测试代码====================//            8//        6      10//       5 7    9  11int main(){BinaryTreeNode* pNode8 = CreateBinaryTreeNode(8);BinaryTreeNode* pNode6 = CreateBinaryTreeNode(6);BinaryTreeNode* pNode10 = CreateBinaryTreeNode(10);BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);BinaryTreeNode* pNode7 = CreateBinaryTreeNode(7);BinaryTreeNode* pNode9 = CreateBinaryTreeNode(9);BinaryTreeNode* pNode11 = CreateBinaryTreeNode(11);ConnectTreeNodes(pNode8, pNode6, pNode10);ConnectTreeNodes(pNode6, pNode5, pNode7);ConnectTreeNodes(pNode10, pNode9, pNode11);//===//PrintTree(pNode8);//===PostorderTraversal(pNode8);cout << endl;//===PostorderTraversal2(pNode8);//==DestroyTree(pNode8);}