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.

给定一个二叉树的先序遍历和中序遍历，构造出一颗二叉树。

二叉树的遍历分为先序遍历、中序遍历、后序遍历、层序遍历。

而通过先序遍历和中序遍历、中序遍历和后序遍历是可以还原该二叉树结构的。

 1     private int getIndexInInorder(int[] inorder, int val) {
 2         for (int i = 0; i < inorder.length; i++)  if (val == inorder[i]) return i;
 3         return -1;
 4     }
 5     private TreeNode build(int[] preorder, int[] inorder, int preIndex,
 6                           int startInIndex, int endInIndex) {
 7         if (endInIndex < startInIndex) return null;
 8         TreeNode node = new TreeNode(preorder[preIndex]);//先序，第一个节点一定是根节点
 9         // the index of current node in inorder
10         int index = getIndexInInorder(inorder,node.val); //获取该节点在中序数组中的位置，进一步结合startInIndex和endInIndex可以获得左右字树的范围
11         int lenL = index - startInIndex;
12         int lenR = endInIndex - index;
13         if (lenL > 0) node.left = build(preorder, inorder, preIndex + 1, startInIndex,
14                 index - 1);
15         if (lenR > 0) node.right = build(preorder, inorder, preIndex + lenL + 1,
16                 index + 1, endInIndex);
17         return node;
18     }
19     
20     public TreeNode buildTree(int[] preorder, int[] inorder) {
21         if (preorder == null || preorder.length == 0) return null;
22         if (inorder == null || inorder.length == 0) return null;
23         if (preorder.length != inorder.length) return null;
24         return build(preorder, inorder, 0, 0, inorder.length - 1);        
25     }