【Construct Binary Tree from Inorder and Postorder Traversal】cpp
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)
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第二次过这道题,沿用了之前construct binary tree的思路,代码一次AC。
/** * 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) { return Solution::build(inorder, 0, inorder.size()-1, postorder, 0, postorder.size()-1); } TreeNode* build( vector<int>& inorder, int bi, int ei, vector<int>& postorder, int bp, int ep) { if ( bi>ei || bp>ep) return NULL; TreeNode* root = new TreeNode(postorder[ep]); int right_range = ei - Solution::findPos(inorder, bi, ei, postorder[ep]); int left_range = ei - bi - right_range; root->left = Solution::build(inorder, bi, ei-right_range-1, postorder, bp, ep-right_range-1); root->right = Solution::build(inorder, bi+left_range+1 , ei, postorder, bp+left_range, ep-1); return root; } int findPos(vector<int>& order, int begin, int end, int val) { for ( int i=begin; i<=end; ++i ) if (order[i]==val) return i; } };