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.

For example, given

inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]
Return the following binary tree:

    3
   / \
  9  20
    /  \
   15   7

难度：medium

题目：给定二叉树中序及后序遍历，构造二叉树，注意二叉树中的结点不重复。

思路；分治。

1. 从后序遍历数组中由后向前取结点r
2. 在中序遍历中找到r的位置，并由此结点分成两部分，继续执行1.

Runtime: 4 ms, faster than 68.02% of Java online submissions for Construct Binary Tree from Inorder and Postorder Traversal.

Memory Usage: 37.6 MB, less than 42.45% of Java online submissions for Construct Binary Tree from Inorder and Postorder Traversal.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        if (null == postorder || postorder.length < 1) {
            return null;
        }
        
        return buildTree(inorder, 0, inorder.length - 1, postorder, postorder.length - 1);
    }
    
    public TreeNode buildTree(int[] inorder, int start, int end, int[] postorder, int idx) {
        if (start > end || idx < 0) {
            return null;
        }   
        
        TreeNode root = new TreeNode(postorder[idx]);
        int rIdx = start;
        for (; rIdx <= end; rIdx++) {
            if (inorder[rIdx] == postorder[idx]) {
                break;
            }
        }
        
        root.left = buildTree(inorder, start, rIdx - 1, postorder, idx - (end - rIdx) - 1);
        root.right = buildTree(inorder, rIdx + 1, end, postorder, idx - 1);
        
        return root;
    }
}