Construct Binary Tree from Inorder and Postorder Traversal

LintCode-72.Construct Binary Tree from Inorder and Postorder Traversal

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

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

Given inorder [1,2,3] and postorder [1,3,2], return a tree:

  2
 / \
1   3

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
 
 
public class Solution {
    /**
     *@param inorder : A list of integers that inorder traversal of a tree
     *@param postorder : A list of integers that postorder traversal of a tree
     *@return : Root of a tree
     */
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        if (inorder == null || inorder.length == 0 || 
            postorder == null || postorder.length == 0) {
            return null;
        }
        
        return buildTree(inorder, 0, inorder.length - 1, 
                postorder, 0, postorder.length - 1);
    }
    
    private TreeNode buildTree(int[] inorder, int inStart, int inEnd,
                        int[] postorder, int postStart, int postEnd) {
        if (inStart > inEnd || postStart > postEnd) {
            return null;
        }
        
        int rootVal = postorder[postEnd];
        int rootIndex = 0;
        
        for (int i = inStart; i <= inEnd; i++) {
            if (inorder[i] == rootVal) {
                rootIndex = i;
                break;
            }
        }
        
        int len = rootIndex - inStart;
        TreeNode root = new TreeNode(rootVal);
        root.left = buildTree(inorder, inStart, rootIndex - 1, postorder, postStart, postStart + len - 1);
        root.right = buildTree(inorder, rootIndex + 1, inEnd, postorder, postStart + len, postEnd - 1);
        
        return root;
    }
}

Hope this helps,
Michael