Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.

Solution1: Recursive做法

Example:
          1
     2        3
  4    5    6    7
8  9      10

In order:   8 4 9 2 5 1* 10 6 3 7
Post order: 8 9 4 5 2 10 6 7 3 1*

与105题 http://www.jianshu.com/p/50598ba9e342 类似

思路：
Postorder的最后一位是根节点，顺序:左树，右树，根节点；Inorder顺序：左树，根节点，右树。
利用此特性，每次在每一段postorder序列中，得到最后一位元素，在inorder序列中查找此元素位置pos，inorder序列中pos的前、后部分a(8 4 9 2 5)、b(10 6 3 7)分别作为下次递归的inorder序列input；
并利用a、b部分的长度/位置在postorder序列中找到相应部分c(8 9 4 5 2)、d(6 7 3 1)，作为下次递归的postorder序列input。
重复此过程(Top-down Recursion)。
在inorder查找的过程利用hashmap，故
Time Complexity: O(N) Space Complexity: O(N)

1.b change End to 元素边界

Solution2: Iterative做法

思路：

Solution1.a Code：

class Solution1 {
    private HashMap<Integer, Integer> inorder_map = new HashMap<Integer, Integer>();
    
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        if (inorder == null || postorder == null || inorder.length != postorder.length)
            return null;

        // build hashmap from Inorder array
        for (int i = 0; i < inorder.length; ++i) {
            inorder_map.put(inorder[i], i);
        }

        // recursively build the tree
        return build(inorder, 0, inorder.length - 1, postorder, 0, postorder.length - 1);
    }

    public TreeNode build(int[] inorder, int in_start, int in_end, int[] postorder, int post_start, int post_end) {
        if(in_start > in_end || post_start > post_end)
            return null;

        int root_value = postorder[post_end];
        TreeNode root = new TreeNode(root_value);

        int pos_inorder = inorder_map.get(root_value);
        int post_left_length = pos_inorder - in_start;
        root.left = build(inorder, in_start, pos_inorder - 1, postorder, post_start, post_start + post_left_length - 1);
        root.right = build(inorder, pos_inorder + 1, in_end, postorder, post_start + post_left_length, post_end - 1);

        return root;

    }
}

Solution1.b Code：
change End to 元素边界

class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        if(inorder == null || postorder == null || inorder.length == 0 || postorder.length == 0) return null;
        
        Map<Integer, Integer> in_map = new HashMap<>();
        for(int i = 0; i < inorder.length; i++) {
            in_map.put(inorder[i], i);
        }
        
        return dfsBuild(in_map, inorder, 0, inorder.length, postorder, 0, postorder.length);
        
    }
    
    private TreeNode dfsBuild(Map<Integer, Integer> in_map, int[] inorder, int in_s, int in_e, int[] postorder, int post_s, int post_e) {
        if(in_s == in_e || post_s == post_e || post_e - 1 < 0) return null;
        
        int cur_val = postorder[post_e - 1];
        TreeNode root = new TreeNode(cur_val);
        int in_index = in_map.get(cur_val);
        int rela_in_index = in_index - in_s;
        
        root.left = dfsBuild(in_map, inorder, in_s, in_index, postorder, post_s, post_s + rela_in_index);
        root.right = dfsBuild(in_map, inorder, in_index + 1, in_e, postorder, post_s + rela_in_index, post_e - 1);
        
        return root;
    }
}
``