学步园

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

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

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
		// Start typing your Java solution below
		// DO NOT write main() function
		HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
		for (int i = 0; i < inorder.length; i++) {
			map.put(inorder[i], i);
		}
		return buildRec(map, inorder, 0, inorder.length - 1, postorder, 0,
				postorder.length - 1);
	}

	private TreeNode buildRec(HashMap<Integer, Integer> map, int[] inorder,
			int is, int ie, int[] postorder, int ps, int pe) {
		if (is > ie)
			return null;
		TreeNode root = new TreeNode(postorder[pe]);
		if (is == ie)
			return root;
		int i = map.get(postorder[pe]);
		int leftLength = i - is;
		root.left = buildRec(map, inorder, is, i - 1, postorder, ps, ps + i
				- is - 1);
		root.right = buildRec(map, inorder, i + 1, ie, postorder, ps + i - is,
				pe - 1);
		return root;
	}
}