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

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

For example, given

preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]

Return the following binary tree:

    3
   / \
  9  20
    /  \
   15   7

方法1：

/**
 * 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[] preorder, int[] inorder) {   
    return helper(0, 0, inorder.length - 1, preorder, inorder);
  }
   public TreeNode helper(int preStart, int inStart, int inEnd, int[] preorder, int[] inorder)    {
    if (preStart > preorder.length - 1 || inStart > inEnd) {
        return null;
    }
    TreeNode root = new TreeNode(preorder[preStart]);
    int inIndex = 0; 
    for (int i = inStart; i <= inEnd; i++) {
        if (inorder[i] == root.val) {
            inIndex = i;
        }
    }
    root.left = helper(preStart + 1, inStart, inIndex - 1, preorder, inorder);
    root.right = helper(preStart + inIndex - inStart + 1, inIndex + 1,inEnd,preorder,inorder);
    return root;
  }
}
 

時間複雜度：O(n^2)

空間複雜度：O(n)

原始碼github地址：https://github.com/zhangyu345293721/leetcode