Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the
tree.
/** * Definition for binary tree * struct 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) { return build(inorder,0,inorder.size()-1,postorder,0,postorder.size()-1); } TreeNode *build(vector<int> &inorder,int inl,int inr, vector<int> &postorder,int pl,int pr) { if(inl>inr) return NULL; TreeNode *root=new TreeNode(postorder[pr]); int index=inl; while(inorder[index]!=postorder[pr]) index++; root->left=build(inorder,inl,index-1,postorder,pl,pl+index-inl-1); root->right=build(inorder,index+1,inr,postorder,pr-(inr-index),pr-1); } };
Construct Binary Tree from Inorder and Postorder Traversal,布布扣,bubuko.com
Construct Binary Tree from Inorder and Postorder Traversal
原文:http://www.cnblogs.com/erictanghu/p/3759603.html