[LeetCode] Construct Binary Tree from Preorder and Inorder Traversal

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

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

 

方法一：最先想到的就是遞迴，注意low、high的計算，還有初始時NULL情況的處理。。

從inorder中尋找pre的第一個，然後左右遞迴

 1 class Solution 2 { 3     public: 4         TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) 5         { 6             if(preorder.size() == 0 || inorder.size() == 0) 7                 return NULL; 8             return buildTree(preorder, 0, preorder.size()-1, 9                               inorder, 0, inorder.size()-1);10         }     11               12         TreeNode *buildTree(vector<int> &preorder, int low1, int high1,13                             vector<int> &inorder, int low2, int high2)14         {     15             //cout << "==============" <<endl;16             //cout << "low1 = \t" << low1 <<endl;17             //cout << "high1= \t" << high1 <<endl;18             //cout << "low2 = \t" << low2 <<endl;19             //cout << "high2= \t" << high2 <<endl;20               21             TreeNode * p = new TreeNode(preorder[low1]);22             if(low1 == high1)23             {   24                 return p;25             }   26             int index = 0;27             for(index = low2; index < high2; index++)28             {   29                 if(inorder[index] == preorder[low1])30                     break;31             }32             //cout << "index= \t" << index<<endl;33 34             if(index != low2)35                 p->left = buildTree(preorder, low1+1,(low1+1) + (index-1-low2), inorder, low2, index-1);36             if(index != high2)37                 p->right = buildTree(preorder, high1 - (high2-index-1) ,high1, inorder, index+1, high2);38 39             return p;40         }41 } ;

方法二：迭代