二叉树的初始化

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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 treestruct 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：

0 0