[LeetCode]144.Binary Tree Preorder Traversal
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【题目】
Given a binary tree, return the preorder traversal of its nodes' values.
For example:
Given binary tree {1,#,2,3}
,
1 \ 2 / 3
return [1,2,3]
.
Note: Recursive solution is trivial, could you do it iteratively?
【代码一】
/********************************** 日期:2014-10-15* 作者:SJF0115* 题号: 144.Binary Tree Preorder Traversal* 来源:https://oj.leetcode.com/problems/binary-tree-preorder-traversal/* 结果:AC* 来源:LeetCode* 总结:**********************************/#include <iostream>#include <malloc.h>#include <vector>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> v; void PreOrder(TreeNode *root){ if (root == NULL){ return; } v.push_back(root->val); PreOrder(root->left); PreOrder(root->right); } vector<int> preorderTraversal(TreeNode *root) { PreOrder(root); 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.preorderTraversal(root); for(int i = 0;i < v.size();i++){ cout<<v[i]<<endl; }}
【代码二】
非递归实现
class Solution {public: vector<int> preorderTraversal(TreeNode *root) { vector<int> v; stack<TreeNode*> s; TreeNode* p = root; //栈不空或者p不空时循环 while(p != NULL || !s.empty()){ if(p != NULL){ //访问根节点 v.push_back(p->val); //根节点插入栈中,用来访问右子树 s.push(p); //遍历左子树 p = p->left; } else{ //左子树访问完毕,访问右子树 p = s.top(); s.pop(); p = p->right; } } return v; }};
【代码三】
/*------------------------------------------------* 日期:2015-03-25* 作者:SJF0115* 题目: 144.Binary Tree Preorder Traversal* 来源:https://oj.leetcode.com/problems/binary-tree-preorder-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> preorderTraversal(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.push_back(node->val); // 右子树 if(node->right){ s.push(node->right); }//if // 左子树 if(node->left){ s.push(node->left); }//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.preorderTraversal(root); for(int i = 0;i < result.size();++i){ cout<<result[i]<<" "; } cout<<endl; return 0;}
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