[LeetCode]145.Binary Tree Postorder Traversal

来源：互联网发布：网络市场结构特征分析编辑：程序博客网时间：2024/06/10 15:56

【题目】

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?

【代码】

/**********************************   日期：2014-12-07*   作者：SJF0115*   题号: 145.Binary Tree Postorder Traversal*   来源：https://oj.leetcode.com/problems/binary-tree-postorder-traversal/*   结果：AC*   来源：LeetCode*   总结：**********************************/#include <iostream>#include <malloc.h>#include <vector>#include <stack>using namespace std;struct TreeNode {    int val;    TreeNode *left;    TreeNode *right;    TreeNode(int x) : val(x), left(NULL), right(NULL) {}};class Solution {public:    vector<int> postorderTraversal(TreeNode *root) {        vector<int> v;        stack<TreeNode *> stack;        TreeNode *p = root;        TreeNode *q;        do{            //遍历左子树            while(p != NULL){                stack.push(p);                p = p->left;            }            q = NULL;            while(!stack.empty()){                p = stack.top();                stack.pop();                // 右子树是否为空或者已访问过                if(p->right == q){                    v.push_back(p->val);                    //保留访问过的节点                    q = p;                }                else{                    //当前节点不能访问，p节点重新入栈                    stack.push(p);                    //处理右子树                    p = p->right;                    break;                }//if            }//while        }while(!stack.empty());//while        return v;    }};//按先序序列创建二叉树int CreateBTree(TreeNode* &T){    char data;    //按先序次序输入二叉树中结点的值（一个字符），‘#’表示空树    cin>>data;    if(data == '#'){        T = NULL;    }    else{        T = (TreeNode*)malloc(sizeof(TreeNode));        //生成根结点        T->val = data-'0';        //构造左子树        CreateBTree(T->left);        //构造右子树        CreateBTree(T->right);    }    return 0;}int main() {    Solution solution;    TreeNode* root(0);    CreateBTree(root);    vector<int> v = solution.postorderTraversal(root);    for(int i = 0;i < v.size();i++){        cout<<v[i]<<endl;    }}

【代码二】

/*------------------------------------------------*   日期：2015-03-25*   作者：SJF0115*   题目: 145.Binary Tree Postorder Traversal*   来源：https://oj.leetcode.com/problems/binary-tree-postorder-traversal/*   结果：AC*   来源：LeetCode*   博客：------------------------------------------------------*/#include <iostream>#include <stack>#include <vector>using namespace std;// 二叉树节点结构struct TreeNode{    int val;    TreeNode *left;    TreeNode *right;    TreeNode(int x):val(x),left(nullptr),right(nullptr){}};class Solution {public:    vector<int> postorderTraversal(TreeNode *root) {        vector<int> result;        if(root == nullptr){            return result;        }//if        stack<TreeNode*> s;        s.push(root);        TreeNode *node;        while(!s.empty()){            node = s.top();            s.pop();            result.insert(result.begin(),node->val);            // 左子树            if(node->left){                s.push(node->left);            }//if            // 右子树            if(node->right){                s.push(node->right);            }//if        }//while        return result;    }};// 1.创建二叉树void CreateTree(TreeNode* &root){    int val;    //按先序次序输入二叉树中结点的值,‘-1’表示空树    cin>>val;    // 空节点    if(val == -1){        root = nullptr;        return;    }//if    root = new TreeNode(val);    //构造左子树    CreateTree(root->left);    //构造右子树    CreateTree(root->right);}int main() {    freopen("C:\\Users\\Administrator\\Desktop\\c++.txt", "r", stdin);    TreeNode* root = nullptr;    vector<int> result;    // 创建二叉树    CreateTree(root);    Solution solution;    result = solution.postorderTraversal(root);    for(int i = 0;i < result.size();++i){        cout<<result[i]<<" ";    }    cout<<endl;    return 0;}

1 0