leecode 解题总结：106. Construct Binary Tree from Inorder and Postorder Traversal

#include <iostream>#include <stdio.h>#include <vector>using namespace std;/*问题:Given inorder and postorder traversal of a tree, construct the binary tree.Note:You may assume that duplicates do not exist in the tree.分析：这是根据后序和中序构建二叉树。默认不存在重复的元素。前序的特点是根节点在前面中序的特点左根右1】可以每次先寻找后序中从后向前的根节点root，然后在中序中找到该根节点将根节点划分成左子树和右子树2】然后寻找下一个根节点，重复1的步骤假设前序为A[1...n],中序为B[1...n]。那么当前根节点为A的第i个结点i=n,寻找到B中j的位置对应结点为根节点，则有B[1..j-1],B[j+1...n]    注意找到根节点划分后，中序对应左子树的根节点范围在A[2...j]，右子树的根节点范围在A[j+1...n]i=n-1,寻找到B中k的位置对应新的根节点，假设1<=k <= j-1,则将B[1...j-1]继续划分为    B[1...k-1],B[k+1..j-1]当出现B[p...q]时，且p=q,说明此时划分后只有一个结点，后续无需再划分，并根据B[p]来构建一个结点并返回，举例：给定一个前序是        5  4     7     2    6    9    1 3       8前序是:5 4 2 1 3 7 6 9 8中序是:1 2 3 4 5 6 7 8 9后序是:1 3 2 4 6 8 9 7 5输入:9(元素个数)1 2 3 4 5 6 7 8 9(中序)1 3 2 4 6 8 9 7 5(后序)输出:5 4 2 1 3 7 6 9 8(前序)关键:11】可以每次先寻找前序中从后向前的根节点root，然后在中序中找到该根节点将根节点划分成左子树和右子树2】然后寻找下一个根节点，重复1的步骤假设前序为A[1...n],中序为B[1...n]。那么当前根节点为A的第i个结点i=1,寻找到B中j的位置对应结点为根节点，则有B[1..j-1],B[j+1...n]    注意找到根节点划分后，中序对应左子树的根节点范围在A[1...j-1]，右子树的根节点范围在A[j...n-1]i=2,寻找到B中k的位置对应新的根节点，假设1<=k <= j-1,则将B[1...j-1]继续划分为    B[1...k-1],B[k+1..j-1]当出现B[p...q]时，且p=q,说明此时划分后只有一个结点，后续无需再划分，并根据B[p]来构建一个结点2//将中序划分成两部分int leftLen = index - inBeg;//中序左子树半部分，对应后序前半部分TreeNode* leftNode = dfs(postorder , preBeg , preBeg + leftLen - 1 , inorder , inBeg , index - 1 );//中序右子树部分，对应后序后半部分TreeNode* rightNode = dfs(postorder, preBeg + leftLen , preEnd - 1 , inorder , index + 1 , inEnd);*/ struct TreeNode {int val;TreeNode *left;TreeNode *right;TreeNode(int x) : val(x), left(NULL), right(NULL) {}};class Solution {public:TreeNode* dfs(vector<int>& postorder, int preBeg , int preEnd, vector<int>& inorder , int inBeg , int inEnd){if(postorder.empty() || inorder.empty() || (inBeg > inEnd) || (preBeg > preEnd) ){return NULL;}if(inBeg == inEnd){TreeNode* node = new TreeNode(inorder.at(inBeg));return node;}//对结点进行划分，首先获取根节点的值int root = postorder.at(preEnd);//寻找根节点在中序中的位置int index = -1;for(int i = inBeg ; i <= inEnd ; i++){if(root == inorder.at(i)){index = i;break;}}//如果没有寻找到根节点if(index == -1){return NULL;}TreeNode* rootNode = new TreeNode(root);//将中序划分成两部分int leftLen = index - inBeg;//中序左子树半部分，对应后序前半部分TreeNode* leftNode = dfs(postorder , preBeg , preBeg + leftLen - 1 , inorder , inBeg , index - 1 );//中序右子树部分，对应后序后半部分TreeNode* rightNode = dfs(postorder, preBeg + leftLen , preEnd - 1 , inorder , index + 1 , inEnd);if(leftNode){rootNode->left = leftNode;}if(rightNode){rootNode->right = rightNode;}return rootNode;}    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {if(postorder.empty() || inorder.empty() || ( postorder.size() != inorder.size() ) ){return NULL;}        TreeNode* root = dfs(postorder , 0 , postorder.size() - 1 , inorder , 0 , inorder.size() - 1);return root;            }};void print(vector<int>& result){if(result.empty()){cout << "no result" << endl;return;}int size = result.size();for(int i = 0 ; i < size ; i++){cout << result.at(i) << " " ;}cout << endl;}//前序遍历树void preorderVisit(TreeNode* root , vector<int>& result){if(!root){return ;}if(root){result.push_back(root->val);}preorderVisit(root->left, result);preorderVisit(root->right , result);}void deleteTree(TreeNode* root){if(!root){return;}if(NULL == root->left && NULL == root->right){delete(root);root = NULL;}if(root){deleteTree(root->left);deleteTree(root->right);}}void process(){ vector<int> postorder; vector<int> inorder; int value; int num; Solution solution; vector<int> result; while(cin >> num ) { postorder.clear(); inorder.clear(); for(int i = 0 ; i < num ; i++) { cin >> value; inorder.push_back(value); } for(int i = 0 ; i < num ; i++) { cin >> value; postorder.push_back(value); } //接下来构建树 TreeNode* root = solution.buildTree(inorder , postorder); preorderVisit(root , result); print(result); deleteTree(root); }}int main(int argc , char* argv[]){process();getchar();return 0;}