【LeetCode】-Construct Binary Tree from Preorder and Inorder Traversal

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

Note:

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

/** * 根据先序数组和中序数组，建立二叉树 * 解题思路： * 1.取先序数组的第一个元素作为根元素； * 2.找到根元素在中序数组中的位置index，小于index为左子树，大于index为右子树； * 3.分别对左右子树重复1.2操作。 *  * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {        private int getIndex( int[] nums, int element, int start, int end ){for( int i=start; i<=end; i++ ){if( element==nums[i] )return i;}return -1;}private TreeNode createTree( int[] pre_arr,int pre_start,int[] in_arr,int in_start,int length){if( length<=0 ){return null;}//取先序数组的第一个元素，作为跟节点TreeNode root = new TreeNode(pre_arr[pre_start]);//计算根节点元素在先序数组中的下标int index = getIndex( in_arr, pre_arr[pre_start], in_start, in_start+length-1 );int len = index - in_start;//对左右子树递归求解root.left = createTree( pre_arr, pre_start+1, in_arr, in_start, len);root.right = createTree( pre_arr, pre_start+len+1, in_arr, index+1, length-len-1);return root;}    public TreeNode buildTree(int[] preorder, int[] inorder) {        if( preorder.length!=inorder.length )return null;return createTree( preorder, 0, inorder, 0, inorder.length );            }}