leetcode 重建二叉樹

1、Construct Binary Tree from Preorder and Inorder Traversal
鏈接：https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/
思路：根據先序遍歷和中序遍歷，重建樹。遞歸思想，因爲樹節點不重複，先序的第一個爲根，在中序的數組中找到，中序左邊爲左子樹，右邊爲右子樹。

    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
鏈接：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;
    }