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.

见剑指offer重建二叉树

class Solution {

    public TreeNode buildTree(int[] preorder, int[] inorder) {

        return helper(0, 0, inorder.length - 1, preorder, inorder);

    }

    

    public TreeNode helper(int preStart, int inStart, int inEnd, int[] preorder, int[] inorder) {

        if (preStart > preorder.length - 1 || inStart > inEnd)

            return null;

        TreeNode root = new TreeNode(preorder[preStart]);

        int inIndex = 0;

        for (int i = inStart; i <= inEnd; i++) {

            if (inorder[i] == root.val)

                inIndex = i;

        }

        root.left = helper(preStart + 1, inStart, inIndex - 1, preorder, inorder);

        root.right = helper(preStart + inIndex - inStart + 1, inIndex + 1, inEnd, preorder, inorder);

        return root;

    }

}