给定先序和后续，构造出一颗二叉树并输出中序序列

最新推荐文章于 2024-01-08 16:50:05 发布

Southan97

最新推荐文章于 2024-01-08 16:50:05 发布

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分类专栏： Data Structure 文章标签：二叉树构造树

本文链接：https://blog.youkuaiyun.com/Mrx_Nh/article/details/72455302

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本文介绍了一种根据前序遍历和中序遍历序列构建二叉树的方法，并提供了详细的算法步骤及C++实现代码。该算法首先从预定义的前序序列中选取元素并递增前序索引，创建一个新节点，接着通过查找中序序列中的对应位置来递归地构建左右子树。

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Algorithm: buildTree()
1) Pick an element from Preorder. Increment a Preorder Index Variable (preIndex in below code) to pick next element in next recursive call.
2) Create a new tree node tNode with the data as picked element.
3) Find the picked element’s index in Inorder. Let the index be inIndex.
4) Call buildTree for elements before inIndex and make the built tree as left subtree of tNode.
5) Call buildTree for elements after inIndex and make the built tree as right subtree of tNode.

6) return tNode.

#include <bits/stdc++.h>
 
using namespace std;

struct Node {
	Node(){}
	Node(char data) : data(data) {
	 	left = right = NULL;
	}
 	char data;
 	Node *left, *right;
};

int Search(char *arr, int inst, int inend, char val) {
	for(int i = inst; i <= inend; ++i) {
	 	if(arr[i] == val) {
	 	 	return i;
	 	}
 	}
 	return -1;
}

Node* build(char *in, char *pre, int inst, int inend) {
	static int preIndex = 0;   // just define when first call build, like a overall variable
	Node *tNode = new Node(pre[preIndex++]);
    if(inst > inend) {
     	return NULL;
    }
    if(inst == inend) {
     	return tNode;
    }
    int index = Search(in, inst, inend, tNode->data);
    tNode->left = build(in, pre, inst, index-1);
    tNode->right = build(in, pre, index+1, inend);
    return tNode; 	
}

void inOrder(Node *t) {
 	if(t != NULL) {
 	 	inOrder(t->left);
 	 	cout << t->data << " ";
 	 	inOrder(t->right);
 	}
}

int main(){
    ios::sync_with_stdio(false);
    char in[] = {'D', 'B', 'E', 'A', 'F', 'C'};
    char pre[] = {'A', 'B', 'D', 'E', 'C', 'F'};
    int length = sizeof(in) / sizeof(in[0]);
    Node *root = build(in, pre, 0, length - 1);
    cout << "Inorder traversal of the binary tree is: " << endl;
    inOrder(root);
    return 0;   
}