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

Note:

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

Solution:

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    TreeNode *build(vector
     
       &preorder, vector
      
        &inorder, int pre_start, int pre_end, int in_start, int in_end){    if(pre_start > pre_end || in_start > in_end)        return NULL;    TreeNode *curRoot = new TreeNode(preorder[pre_start]);    int rootIndex = -1;    for(int i = in_start;i <= in_end;i++)    {        if(inorder[i] == preorder[pre_start])        {            rootIndex = i;            break;        }    }    if(rootIndex == -1) return NULL;    int leftNum = rootIndex - in_start;    curRoot -> left = build(preorder, inorder, pre_start + 1, pre_start + leftNum, in_start, rootIndex - 1);    curRoot -> right = build(preorder, inorder, pre_start + leftNum + 1, pre_end, rootIndex + 1, in_end);    return curRoot;}TreeNode *buildTree(vector
       
         &preorder, vector
        
          &inorder) {        return build(preorder, inorder, 0, preorder.size() - 1, 0, inorder.size() - 1);    }};