【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;}
};