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题目:
Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
代码:
/** * Definition for a binary tree node. * 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) { if ( inorder.size()==0 || postorder.size()==0 ) return NULL; return Solution::buildTreeIP(inorder, 0, inorder.size()-1, postorder, 0, postorder.size()-1); } static TreeNode* buildTreeIP( vector<int>& inorder, int bI, int eI, vector<int>& postorder, int bP, int eP ) { if ( bI > eI ) return NULL; TreeNode *root = new TreeNode(postorder[eP]); int rootPosInorder = bI; for ( int i = bI; i <= eI; ++i ) { if ( inorder[i]==root->val ) { rootPosInorder=i; break; } } int leftSize = rootPosInorder - bI; int rightSize = eI - rootPosInorder; root->left = Solution::buildTreeIP(inorder, bI, rootPosInorder-1, postorder, bP, bP+leftSize-1); root->right = Solution::buildTreeIP(inorder, rootPosInorder+1, eI, postorder, eP-rightSize, eP-1); return root; } };
tips:
思路跟Preorder & Inorder一样。
这里要注意:
1. 算左子树和右子树长度时,要在inorder里面算
2. 左子树和右子树长度可能一样,也可能不一样;因此在计算root->left和root->right的时候,要注意如何切vector下标(之前一直当成左右树长度一样,debug了一段时间才AC)
【Construct Binary Tree from Inorder and Postorder Traversal】cpp
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原文地址:http://www.cnblogs.com/xbf9xbf/p/4507596.html