Binary Tree Postorder Traversal

Description:

Given a binary tree, return the postorder traversal of its nodes’ values.
For example: Given binary tree {1,#,2,3},

return [3,2,1].
Note: Recursive solution is trivial, could you do it iteratively?

分析：
使用栈

代码：

#include <iostream>#include <stack>#include <vector>#define Elementype charusing namespace std;typedef struct TreeNode //树结点{    Elementype  val;    TreeNode *left;    TreeNode *right;    TreeNode(Elementype x) : val(x), left(nullptr), right(nullptr) {}} *Tree;int index = 0;  //全局索引变量//使用先序遍历创建创建二叉树//左子树或右子树不存在用"#"表示void MakeBinaryTree(Tree &T, Elementype value[]){    Elementype c = value[index++];    if (c == '#')        T = nullptr;    else    {        T = new TreeNode(c);        MakeBinaryTree(T->left, value);        MakeBinaryTree(T->right, value);    }}//非递归先序遍历//使用栈vector<Elementype>  BinaryTreepostorderTraversal(Tree root){    vector<Elementype> result;    stack<Tree> sk;    Tree p = root;    Tree pre = nullptr;    while (!sk.empty() || p != nullptr)    {        while (p) //沿左子树一直往下走        {            sk.push(p);            p = p->left;        }        p = sk.top();        //如果p没有右孩子或者其右孩子刚刚被访问过        if (p->right == nullptr || p->right == pre)        {            result.push_back(p->val);            sk.pop();            pre = p;            p = nullptr;        }        else        {            p = p->right;        }    }    return result;}int main(){    Tree T = nullptr;    Elementype data[7] = {'1','#','2','3','#','#','#'}; //注意，每个结点都要有值    vector<Elementype> result;    MakeBinaryTree(T, data);    result = BinaryTreepostorderTraversal(T);    for (auto &s : result)        cout << s << " ";    cout << endl;    system("pause");    return 0;}

测试：