leetcode 重建二叉树

1、Construct Binary Tree from Preorder and Inorder Traversal
链接：https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/
思路：根据先序遍历和中序遍历，重建树。递归思想，因为树节点不重复，先序的第一个为根，在中序的数组中找到，中序左边为左子树，右边为右子树。

    public TreeNode buildTree(int[] preorder, int[] inorder) {        int size = preorder.length;        if(size == 0)            return null;        return help(preorder, inorder, 0, size - 1, 0, size - 1);    }    public TreeNode help(int[] preorder, int[] inorder, int pleft, int pright, int ileft, int iright) {        if(pleft > pright || ileft > iright)            return null;        TreeNode root = new TreeNode(preorder[pleft]);        int i;        for(i = ileft; i <= iright; i++){            if(inorder[i] == preorder[pleft])                break;        }        root.left = help(preorder, inorder,pleft + 1, pleft + i - ileft, ileft, i - 1);        root.right = help(preorder, inorder,pleft + i - ileft + 1, pright, i + 1, iright);        return root;    }

2、Construct Binary Tree from Inorder and Postorder Traversal
链接：https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/
思路：同上

    public TreeNode buildTree(int[] inorder, int[] postorder) {        int size = inorder.length;        if(size == 0)            return null;        return help(inorder, postorder, 0, size - 1, 0, size - 1);    }    public TreeNode help(int[] inorder, int[] postorder,  int ileft, int iright, int pleft, int pright) {        if(pleft > pright || ileft > iright)            return null;        TreeNode root = new TreeNode(postorder[pright]);        int i;        for(i = ileft; i <= iright; i++){            if(inorder[i] == postorder[pright])                break;        }        root.left = help(inorder, postorder,ileft, i - 1, pleft, pleft + i - ileft - 1);        root.right = help(inorder, postorder,i + 1, iright, pleft + i - ileft, pright - 1);        return root;    }