LeetCode:Construct Binary Tree from Preorder and Inorder Traversal

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#leetcode #前序遍历和中序遍历输出后序遍历

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本文介绍了一种通过前序和中序遍历序列构建二叉树的方法，详细解释了构建过程并提供了一个示例程序。

题目：

Given preorder and inorder traversal of a tree, construct the binary tree.

代码如下：

/*根据前序遍历的序列和中序遍历的序列求出整个树*/
#include <iostream>
#include <exception>
#include <vector>
#include <stdio.h>
using namespace std;
struct TreeNode
{
	TreeNode* left;
	TreeNode* right;
	int val;
	TreeNode(int x):val(x),left(NULL),right(NULL){}
};

class Solution {
public:
    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
		int prelength = preorder.size();
		int inlength = inorder.size();
		vector<int>::iterator itr = preorder.begin();
		TreeNode *root;
		if(prelength == inlength)
		{
			 if(prelength == 0)
		        return NULL;
			if(prelength == 1 && preorder[0]==inorder[0]){
				root = new TreeNode(0);
				root->val = preorder[0];
				return root;
			}
		}
		else
			throw exception();

		root = buildNode(preorder, inorder, 0,prelength-1,0,inlength-1);
		return root;

    }
	void PrintTree(TreeNode *node)
	{
		if(!node)
			return;
		else
		{
			if(node ->left)
				PrintTree(node->left);
			if(node ->right)
				PrintTree(node->right);
			cout << node -> val <<'\t';
		}
	}
	TreeNode *buildNode(vector<int> &preorder,vector<int> &inorder,int pre_start,int pre_end,int in_start,int in_end)
	{
		TreeNode *root = new TreeNode(0);
		//在前序遍历中首先找到根节点
		root->val = preorder[pre_start];
		//再在中序遍历中找到根节点
		
		int root_index = 0;
		for(int i = in_start; i <= in_end; i++)
		{
			if(inorder[i] == root->val)
			{
				root_index = i;
				if(root_index == in_start)
					root->left = NULL;
				else
				{
					int length = root_index - in_start;
					root->left = buildNode(preorder,inorder,pre_start+1,pre_start+length,in_start,root_index-1);
				}
				if(root_index == in_end)
					root->right = NULL;
				else
				{
					int length = in_end - root_index;
					root->right = buildNode(preorder,inorder,pre_end-length+1,pre_end,root_index+1,in_end);
				}
				return root;
			}
		}
	}
};
void Test1()
{
	const int length = 8;
	int pre[length] = {1,2,4,7,3,5,6,8};
	int in[length] = {4,7,2,1,5,3,8,6};
	vector<int> preorder,inorder;
	for(int i = 0; i < length; i++)
	{
		preorder.push_back(pre[i]);
		inorder.push_back(in[i]);
	}
	Solution sol;
	TreeNode *node = NULL;
	node = sol.buildTree(preorder,inorder);
	cout << "The post_order is :";
	sol.PrintTree(node);
	cout << endl;
	return;
}
int main()
{
	Test1();
	system("PAUSE");
	return 0;
}

运行结果：