leetcode106. Construct Binary Tree from Inorder and Postorder Traversal

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.

解法

后序遍历的最后一个元素为根节点，在中序遍历中找到该根节点，中序遍历中，该元素的左边为左子树，右边为右子树。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
          if ((inorder == null || inorder.length == 0) && (postorder == null || postorder.length == 0)) {
            return null;
        }

        return helper(inorder, 0, inorder.length - 1, postorder, 0, postorder.length - 1);
    }
    private TreeNode helper(int[] inorder, int inStart, int inEnd, int[] postorder, int postStart, int postEnd) {
        if (inStart > inEnd || postStart > postEnd) {
            return null;
        }

        TreeNode root = new TreeNode(postorder[postEnd]);
        int inIndex = 0;
        for (int i = inStart; i <= inEnd; i++) {
            if (inorder[i] == root.val) {
                inIndex = i;
                break;
            }
        }

        root.left = helper(inorder, inStart, inIndex - 1, postorder, postStart, postStart + inIndex - inStart - 1);
        root.right = helper(inorder, inIndex + 1, inEnd, postorder, postStart + inIndex - inStart, postEnd - 1);

        return root;
    }
}