105. Construct Binary Tree from Preorder and Inorder Traversal

原题

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

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

代码实现

the most important function in solving this issue is

private TreeNode bulidTree(int preStart, int inStart, int inEnd) ;

Plus, preStart index in preorder is the root index, which is also the separate point in inorder and it’s left is left subtree and right is right subtree.

public class Solution {    private  int[] _preorder; //preorder array    private  int[] _inorder; //inorder array    public TreeNode BuildTree(int[] preorder, int[] inorder)     {        _preorder = preorder;        _inorder = inorder;        //if(_preorder.Length==0 || inorder.Length==0) return null;        //if(prorder.Length!=inorder.Length) return null;        return bulidTree(0,0,_inorder.Length-1);    }    //preStart: index of root of tree    //inStart: of inorder tree, inStart ~ root index - 1 -> left tree    //EndStart:of inorder tree, root index + 1 ~ inEnd -> right tree    private TreeNode  bulidTree(int preStart, int inStart, int inEnd)     {        if (preStart > _preorder.Length - 1 || inStart > inEnd)         {            return null;        }        TreeNode root = new TreeNode(_preorder[preStart]);         // find the index of current root in inorder.         int inIndex = curRootIndex(inStart, inEnd, root);         root.left = bulidTree(preStart + 1, inStart, inIndex - 1);        //right subtree begins position in preorder        preStart += inIndex - inStart + 1;         root.right = bulidTree(preStart, inIndex + 1, inEnd);        return root;    }      private int curRootIndex(int inStart, int inEnd, TreeNode root)      {           for (int i = inStart; i <= inEnd; i++)            {               if (_inorder[i] == root.val)                 {                   return i;                }            }          return -1;        }  }

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