106.Construct Binary Tree from Inorder and Postorder Traversal

题目链接

题目大意：根据中序和后序构建二叉树。

法一：与105类似，只是这里是根据后序来确定根节点。代码如下（耗时15ms）：

    public TreeNode buildTree(int[] inorder, int[] postorder) {
        if(postorder.length == 0 || inorder.length == 0) {
            return null;
        }
        return dfs(inorder, postorder, 0, inorder.length - 1, 0, postorder.length - 1);
    }
    private TreeNode dfs(int[] inorder, int[] postorder, int inL, int inR, int postL, int postR) {
        TreeNode root = new TreeNode(postorder[postR]);
        int rootIndex = inL;
        while(inorder[rootIndex] != postorder[postR]) {
            rootIndex++;
        }
        int leftLen = rootIndex - inL;
        int rightLen = inR - rootIndex;
        if(leftLen != 0) {
            root.left = dfs(inorder, postorder, inL, rootIndex - 1, postL, postL + leftLen - 1);
        }
        else {
            root.left = null;
        }
        if(rightLen != 0) {
            root.right = dfs(inorder, postorder, rootIndex + 1, inR, postL + leftLen, postR - 1);
        }
        else {
            root.right = null;
        }
        return root;
    }