105. Construct Binary Tree from Preorder and Inorder Traversal

problem description

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

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

For example, given

preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]

Return the following binary tree:

    3
   / \
  9  20
    /  \
   15   7

algorithm thought

前序遍历和中序遍历转二叉树，首先利用前序遍历第一个值就是当前根节点的性质。找到根节点，只有去中序遍历中找到根节点对应的位置，位置左边就是左子树右边是右子树。然后对于左右子树，得到他们的大小，去前序遍历中找到他们对应的位置。然后对于左右子树，递归解决。

code

/**
 * 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>& preorder, vector<int>& inorder) {
        return build(preorder,inorder,0,preorder.size(),0,inorder.size());
    }
    TreeNode* build(vector<int>& preorder, vector<int>& inorder,int prelow,int prehigh,int inlow,int inhigh){
        if(prelow==prehigh)
            return NULL; 
        TreeNode* root=new TreeNode(preorder[prelow]);
        int inroot;
        for(int i=inlow;i<inhigh;i++){
            if(inorder[i]==preorder[prelow]){
                inroot=i;
                break;
            }
        }
        int leftsize=inroot-inlow;
        int rightsize=inorder.size()-inroot-1;
        root->left=build(preorder,inorder,prelow+1,prelow+1+leftsize,inlow,inroot);
        root->right=build(preorder,inorder,prelow+1+leftsize,prehigh,inroot+1,inhigh);
        return root;
    }
};

algorithm analysis

每次在中序遍历中找到根节点的位置，一般情况下需要O(n)时间。那总时间可以表达为T(n)=2T(n/2)+O(n)，这个表达式用主定理分析可知为T(n)=O(nlgn)