題目：

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

代碼：

/*** 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一樣。

這里要注意：

1. 算左子樹和右子樹長度時，要在inorder里面算

2. 左子樹和右子樹長度可能一樣，也可能不一樣；因此在計算root->left和root->right的時候，要注意如何切vector下標（之前一直當成左右樹長度一樣，debug了一段時間才AC）

==============================================

第二次過這道題，沿用了之前construct binary tree的思路，代碼一次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;} };

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轉載于:https://www.cnblogs.com/xbf9xbf/p/4507596.html

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