问题:
给定二叉树的初始化数据,怎样动态建立一个二叉树呢?
比如我们给定这样的一组数据:{ 1, 2, 3, 4, 0, 5, 6, 0, 7 }(假设0代表空),则我们构建的二叉树是这样的:
1
/ \
2 3
/ / \
4 5 6
\
7
思路分析:
我们可以使用一个队列,队首出一个元素,队未进两个元素,而这两个元素正好是这个队首元素的左右节点。
参考代码如下:
TreeNode* initBTree(int elements[], int size)
{
if (size < 1)
{
return NULL;
}
//动态申请size大小的指针数组
TreeNode **nodes = new TreeNode*[size];
//将int数据转换为TreeNode节点
for (int i = 0; i < size; i++)
{
if (elements[i] == 0)
{
nodes[i] = NULL;
}
else
{
nodes[i] = new TreeNode(elements[i]);
}
}
queue<TreeNode*> nodeQueue;
nodeQueue.push(nodes[0]);
TreeNode *node;
int index = 1;
while (index < size)
{
node = nodeQueue.front();
nodeQueue.pop();
nodeQueue.push(nodes[index++]);
node->left = nodeQueue.back();
nodeQueue.push(nodes[index++]);
node->right = nodeQueue.back();
}
return nodes[0];
}
下面是一个测试代码,我们可以看看结果:
头文件声明(tree.h):
#pragma once
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
//Definition for binary tree
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
//初始化一个二叉树
TreeNode* initBTree(int elements[], int size);
//树的前序遍历
void preOrder(TreeNode *root, vector<int> &result);
//树的中序遍历
void inOrder(TreeNode *root, vector<int> &result);
//树的后序遍历
void postOrder(TreeNode *root, vector<int> &result);
//vector的遍历
void traverse(vector<int> nums);
源代码实现:
#include "tree.h"
TreeNode* initBTree(int elements[], int size)
{
if (size < 1)
{
return NULL;
}
//动态申请size大小的指针数组
TreeNode **nodes = new TreeNode*[size];
//将int数据转换为TreeNode节点
for (int i = 0; i < size; i++)
{
if (elements[i] == 0)
{
nodes[i] = NULL;
}
else
{
nodes[i] = new TreeNode(elements[i]);
}
}
queue<TreeNode*> nodeQueue;
nodeQueue.push(nodes[0]);
TreeNode *node;
int index = 1;
while (index < size)
{
node = nodeQueue.front();
nodeQueue.pop();
nodeQueue.push(nodes[index++]);
node->left = nodeQueue.back();
nodeQueue.push(nodes[index++]);
node->right = nodeQueue.back();
}
return nodes[0];
}
void preOrder(TreeNode *root, vector<int> &result)
{
if (root)
{
result.push_back(root->val);
preOrder(root->left, result);
preOrder(root->right, result);
}
}
void inOrder(TreeNode *root, vector<int> &result)
{
if (root)
{
inOrder(root->left, result);
result.push_back(root->val);
inOrder(root->right, result);
}
}
void postOrder(TreeNode *root, vector<int> &result)
{
if (root)
{
postOrder(root->left, result);
postOrder(root->right, result);
result.push_back(root->val);
}
}
void traverse(vector<int> nums)
{
vector<int>::size_type size = nums.size();
for (size_t i = 0; i < size; i++)
{
cout << nums[i] << ' ';
}
cout << endl;
}
int main()
{
int nums[] = { 1, 2, 3, 4, 0, 5, 6, 0, 7 };
TreeNode *root = initBTree(nums, 9);
vector<int> preResult;
vector<int> inResult;
vector<int> postResult;
preOrder(root, preResult);
inOrder(root, inResult);
postOrder(root, postResult);
cout << "前序遍历的结果:" << '\n';
traverse(preResult);
cout << "中序遍历的结果:" << '\n';
traverse(inResult);
cout << "后序遍历的结果:" << '\n';
traverse(postResult);
return 0;
}
运行结果如下: