class Solution { public: TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) { int ilen = inorder.size(); int plen = postorder.size(); if (ilen == 0 || plen == 0) return NULL; return dfs(inorder, postorder, 0, ilen, plen - 1, -1); } TreeNode* dfs(vector<int> &inorder, vector<int> &postorder, int is, int ie, int ps, int pe) { if (is >= ie || ps <= pe) return NULL; int rval = postorder[ps]; int iroot= is; while (iroot != ie && inorder[iroot] != rval) iroot++; int left_len = iroot - is; TreeNode* nroot = new TreeNode(rval); nroot->left = dfs(inorder, postorder, is, iroot, pe + left_len, pe); nroot->right= dfs(inorder, postorder, iroot + 1, ie, ps - 1, pe + left_len); return nroot; } };
LeetCode Construct Binary Tree from Inorder and Postorder Traversal,布布扣,bubuko.com
LeetCode Construct Binary Tree from Inorder and Postorder Traversal
原文地址:http://www.cnblogs.com/lailailai/p/3845349.html