一、Description

Return any binary tree that matches the given preorder and postorder traversals.

Values in the traversals pre and post are distinct positive integers.

題目大意：給定二叉樹的前序遍歷序列和後序遍歷序列，返回一個滿足這兩個序列的二叉樹，所有的結點值互不相同。

Example 1:

Input: pre = [1,2,4,5,3,6,7], post = [4,5,2,6,7,3,1]
Output: 
 
[1,2,3,4,5,6,7]

二、Analyzation

對於兩個序列pre[a, b]和post[c, d]，如果我們想通過這兩個序列重構一個二叉樹，顯然pre[a] = post[d]，並且是根結點。

[root][......left......][...right..]  ---pre
[......left......][...right..][root]  ---post

pre[a + 1]是根結點的左子樹的根結點，找到這個結點在post中對應的下標，那麼左子樹的構建就是pre[a + 1, a + index - c + 1]和post[c, index]。

三、Accepted code

class Solution {
    Map<Integer, Integer> map = new HashMap<>();
    public TreeNode constructFromPrePost(int[] pre, int[] post) {
        for (int i = 0; i < post.length; i++) {
            map.put(post[i], i);
        }
        return help(pre, post, 0, pre.length - 1, 0, post.length - 1);
    }
    public TreeNode help(int[] pre, int[] post, int a, int b, int c, int d) {
        if (a > b || c > d) {
            return null;
        }
        TreeNode root = new TreeNode(pre[a]);
        if (a == b || c == d) {
            return root;
        }
        int index = map.get(pre[a + 1]);
        int len = index - c + 1;
        root.left = help(pre, post, a + 1, a + len, c, index);
        root.right = help(pre, post, a + len + 1, b, index + 1, d);
        return root;
    }
}