106. Construct Binary Tree from Inorder and Postorder Tree

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

中序遍历为: 4 2 5 1 6 3 7

后续遍历为: 4 5 2 6 7 3 1

为了清晰表示，我给节点上了颜色，红色是根节点，蓝色为左子树，绿色为右子树。

可以发现的规律是：

1. 中序遍历中根节点是左子树右子树的分割点。

2. 后续遍历的最后一个节点为根节点。

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

class Solution {

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

        if (inorder.length == 0 || postorder.length == 0 || inorder.length != postorder.length)

            return null;

　　  return buildTree(inorder, 0, inorder.length - 1, postorder, 0, postorder.length - 1);        

    }

    

    private TreeNode buildTree(int[] inorder, int inStart, int inEnd, int[] postorder, int postStart, int postEnd) {

        if (inStart > inEnd || postStart > postEnd)

             return null;  //必须要有这个判断，例如输入的两个数组长度都是1，root.left = buildTree() inEnd成为-1，会异常    

        int rootVal = postorder[postEnd];

        int rootIndex = 0;

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

            if (inorder[i] == rootVal) {

                rootIndex = i;

                break;

            }

        }

        

        int len = rootIndex - inStart; //得到根节点后，在中序遍历中，rootIndex - inStart就是左子树长度

        TreeNode root = new TreeNode(rootVal);

        root.left = buildTree(inorder, inStart, rootIndex - 1, postorder, postStart, postStart + len - 1);

        root.right = buildTree(inorder, rootIndex + 1, inEnd, postorder, postStart + len, postEnd - 1);

        return root;

    }

}