105. Construct Binary Tree from Preorder and Inorder Traversal

Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
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根据树的前序和中序遍历得到树的结构，思路为：因为前序遍历从根结点开始，开始从中序遍历中查找，一直查到到根节点，并记录下往后查找的次数count，则左子树在前序中的pree+1到pree+count中，和中序的ins到ins+count-1中，右子树在前序的pres+count+1到pree中和中序的ins+count+1到ine中，在这个过程中采用递归。具体代码如下：

public TreeNode buildTree(int[] preorder, int[] inorder) {
         //根据前序和中序遍历得到树结构。
         return build(preorder, 0, preorder.length-1, inorder, 0, inorder.length-1);
        }
     private static TreeNode build(int[] preorder,int pres,int pree,int[] inorder,int ins,int ine) {
        if(pres>pree) return null;
        int root=preorder[pres];
        int count=0;
        for (int i = ins; i <= ine ; i++) {
            if (inorder[i]!=root) {
                count++;
            }else {
                break;
            }
        }
        TreeNode node=new TreeNode(root);
        node.left=build(preorder, pres+1, pres+count, inorder, ins, ins+count-1);
        node.right=build(preorder, pres+count+1, pree, inorder, ins+count+1, ine);
        return node;
    }