construct-binary-tree-from-inorder-and-postorder-traversal

1、来源：construct-binary-tree-from-inorder-and-postorder-traversal 牛客网

Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.

2、思路，通过中序和后序来构建树：

• 后序的最后一个元素为当前根结点root；
• 在中序中，这个root左边为左子树，右边为右子树，递归就好了

3、代码：

public TreeNode buildTree(int[] inorder, int[] postorder) {        if(inorder == null || inorder.length == 0)            return null;        TreeNode root = biuld(inorder, 0, inorder.length - 1, postorder, 0, postorder.length - 1);        return root;    }    private TreeNode biuld( int[] inorder, int start1, int end1,            int[] postorder, int start2, int end2) {        int root_val = postorder[end2];        TreeNode root = new TreeNode(root_val);        int i;        for(i = start1; i < end1; i++){            if(root_val == inorder[i])                break;        }        if(i - start1 > 0 ){            root.left = biuld( inorder, start1, i - 1, postorder, start2, start2 + i - 1 - start1);        }        if(end1 - i > 0 ){            root.right = biuld( inorder, i + 1, end1, postorder, end2 - end1 + i, end2 - 1);        }        return root;    }