leecode 解题总结:106. Construct Binary Tree from Inorder and Postorder Traversal
2017-02-15 01:37
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#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: 1 1】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; }
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