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Construct Binary Tree from Inorder and Postorder Traversal
Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
C++实现代码：
#include<iostream>#include<new>#include<vector>using namespace std;//Definition for binary treestruct TreeNode{ int val; TreeNode *left; TreeNode *right; TreeNode(int x) : val(x), left(NULL), right(NULL) {}};class Solution {public: TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) { TreeNode *root=NULL; root=build(inorder.begin(),inorder.end(),postorder.begin(),postorder.end()); return root; } TreeNode *build(vector<int>::iterator ibegin,vector<int>::iterator iend,vector<int>::iterator pbegin,vector<int>::iterator pend) { TreeNode *root=NULL; if(pbegin==pend||ibegin==iend) return NULL; auto it=ibegin; auto tmp=pend-1; while(it!=iend) { if(*it==*tmp) break; it++; } root=new TreeNode(*it); int len=it-ibegin; root->left=build(ibegin,it,pbegin,pbegin+len); root->right=build(it+1,iend,pbegin+len,pend-1); return root; } void preorder(TreeNode *root) { if(root) { cout<<root->val<<" "; preorder(root->left); preorder(root->right); } } void inorder(TreeNode *root) { if(root) { inorder(root->left); cout<<root->val<<" "; inorder(root->right); } } void postorder(TreeNode *root) { if(root) { postorder(root->left); postorder(root->right); cout<<root->val<<" "; } }};int main(){ vector<int> posorder={4,5,2,6,7,3,1}; vector<int> inorder={4,2,5,1,6,3,7}; Solution s; TreeNode *root=s.buildTree(inorder,posorder); s.preorder(root); cout<<endl; s.inorder(root); cout<<endl; s.postorder(root); cout<<endl;}
运行结果：
Construct Binary Tree from Inorder and Postorder Traversal
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