您的位置:首页 > 职场人生

面试过程中手撕代码之二叉树

2017-08-31 11:20 197 查看

二叉树的个人总结

由于本人目前在找算法工程师方向的工作,在面试过程中,经常会被问到非递归方法遍历的二叉树,二叉搜索树,二叉树的高度等一系列方法,本人写了c++代码供自己复习使用,希望也能够帮到各位小伙伴们

代码块

//
//  tree.cpp
//  myTest
//
//  Created by lixiaoxue on 2017/8/28.
//  Copyright © 2017年 lixiaoxue. All rights reserved.
//

#include <stdio.h>
#include <iostream>
#include <queue>
#include <vector>
#include <algorithm>
#include <math.h>
#include <numeric>
#include <stack>
using namespace std;
struct TreeNode{
public:
int val;
TreeNode* left;
TreeNode* right;
TreeNode(int a): val(a),left(nullptr),right(nullptr){}
};
//递归先序遍历
void pre_order(vector<int>& out,TreeNode* root){
if(!root) return;
out.push_back(root->val);
pre_order(out, root->left);
pre_order(out, root->right);
}
//递归中序遍历
void mid_order(vector<int>& out,TreeNode* root){
if(!root) return;
mid_order(out, root->left);
out.push_back(root->val);
mid_order(out, root->right);
}
//递归的后序遍历
void after_order(vector<int>& out,TreeNode* root){
if(!root) return;
after_order(out, root->left);
after_order(out, root->right);
out.push_back(root->val);
}

//非递归先序遍历2
vector<int> pre_order_non2(TreeNode* root){
if(!root) return vector<int>();
vector<int> number;
stack<TreeNode*> s;
TreeNode* cur = root;
while ((!s.empty()) || !cur) {
while (cur) {
number.push_back(cur->val);
s.push(cur);
cur = cur->left;
}
if(!s.empty()){
cur = s.top()->right;
s.pop();
}
}
return number;
}
//先序遍历3
vector<int> pre_order_non3(TreeNode* root){
if(!root) return vector<int>();
stack<TreeNode*> s;
vector<int> number;
while (root) {
number.push_back(root->val);
s.push(root);
root = root->left;
}
while (!s.empty()) {
TreeNode* temp = s.top()->right;
s.pop();
while (temp) {
number.push_back(temp->val);
s.push(temp);
temp = temp->left;
}
}
return number;
}
//中序遍历非递归
vector<int> mid_order_non(TreeNode* root){
if(!root) return vector
b78f
<int>();
stack<TreeNode*> s;
vector<int> number;
TreeNode* cur = root;
while ((!s.empty()) || cur) {
while (cur) {
s.push(cur);
cur = cur->left;
}
if (!s.empty()) {
TreeNode* temp = s.top();
s.pop();
number.push_back(temp->val);
cur = temp->right;
}
}
return number;
}
//中序遍历2
vector<int> mid_order_non2(TreeNode* root){
if(!root) return vector<int>();
stack<TreeNode*> s;
vector<int> number;
TreeNode* cur = root->left;
while ((!s.empty())||cur) {
while (cur) {
s.push(cur);
cur = cur->left;
}
cur = s.top();
s.pop();
number.push_back(cur->val);
cur = cur->right;
}
return number;
}

//后序遍历
vector<int> after_order_non(TreeNode* root){
if(!root) return vector<int>();
vector<int> number;
stack<TreeNode*> s;
TreeNode* cur = root;
TreeNode* last = nullptr;
while ((!s.empty()) || cur) {
while (cur) {
s.push(cur);
cur = cur->left;
}
cur = s.top();
if((!cur->right) || cur->right == last){
number.push_back(cur -> val);
last = cur;
s.pop();
cur = s.top();
}else
cur = cur->right;
}
return number;
}

void PostOrder_Nonrecursive(TreeNode* root)  // 后序遍历的非递归     双栈法
{
stack<TreeNode*> s1 , s2;
TreeNode* curr ;           // 指向当前要检查的节点
s1.push(root);
while(!s1.empty())  // 栈空时结束
{
curr = s1.top();
s1.pop();
s2.push(curr);
if(curr->left)
s1.push(curr->left);
if(curr->right)
s1.push(curr->right);
}
while(!s2.empty())
{
printf("%c ", s2.top()->val);
s2.pop();
}
}

//求二叉树非遍历情况下的平均深度
int findDepth(TreeNode* root){
if(!root) return 0;
root->val = 1;
vector<int> height;
vector<TreeNode*> tree{root};
while (tree.size()) {
root = tree.back();
tree.pop_back();
if(!root->left && !root->right)
height.push_back(root->val);
if(root->left){
tree.push_back(root->left);
root->left->val = root->val+1;
}
if(root->right){
tree.push_back(root->right);
root->right->val = root->val + 1;
}
}
int sum = accumulate(height.begin(), height.end(), 0);
cout << sum << "  " << height.size() << endl;
return (float)sum/(float)height.size();
}
//求二叉树非遍历情况下的平均深度
int tree_depth(TreeNode* root){
if(!root) return 0;
vector<TreeNode*> temp{root};
int h =0;
vector<int> height;
while (temp.size()) {
h+= 1;
vector<TreeNode*> nei;
for(TreeNode* it:temp){
if(!it->left && !it->right)
{height.push_back(h);continue;}
if(it->left)
nei.push_back(it->left);
if (it->right) {
nei.push_back(it->right);
}
}
temp = nei;
}
int sum= accumulate(height.begin(), height.end(), 0);
cout << sum << " " << height.size() << endl;
return sum/height.size();
}
int main(){
TreeNode* root = new TreeNode(6);
TreeNode* left1 = new TreeNode(6);
TreeNode* right1 = new TreeNode(6);
TreeNode* left12 = new TreeNode(6);
TreeNode* left123 = new TreeNode(6);
TreeNode* left1234 = new TreeNode(6);
root->left = left1;
left1->left = left12;
left12->left = left123;
left123->right = left1234;
root->right = right1;
int zhi = tree_depth(root);
cout << zhi << endl;
return 0;
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: