106. Construct Binary Tree from Inorder and Postorder Traversal

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

Note:

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

第二次刷 做出来了 感觉自己写的比答案简洁

    private Map<Integer, Integer> map;
    
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        map = new HashMap<>();
        for (int i=0; i<inorder.length; i++) {
            map.put(inorder[i], i);
        }
        return build(0, inorder.length-1, postorder, 0, postorder.length-1);
    }
    
    private TreeNode build(int is, int ie, int[] postorder, int ps, int pe) {
        if (ie<is || pe<ps) return null;
        int val = postorder[pe];
        TreeNode root = new TreeNode(postorder[pe]);
        int index = map.get(val);
        root.left = build(is, index-1, postorder, ps, ps+index-1-is);
        root.right = build(index+1, ie, postorder, ps+index-is, pe-1);
        return root;
    }