从前序与中序遍历序列构造二叉树（中等）

先从前序遍历列表取出第一个元素，这个元素就是根节点，然后从中序遍历中找到这个根节点，节点左侧就是该节点的左子树的节点集合，右侧就是该节点的右侧节点集合，然后递归构建左右子树。

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {private Map<Integer, Integer> indexMap;public TreeNode myBuildTree(int[] preorder,int[] inorder,int preorder_left,int preorder_right,int inorder_left,int inorder_right){if(preorder_left>preorder_right){return null;}int preorder_root=preorder_left;int inorder_root=indexMap.get(preorder[preorder_root]);TreeNode root=new TreeNode(preorder[preorder_root]);int size_left_subtree=inorder_root-inorder_left;root.left=myBuildTree(preorder,inorder,preorder_left+1,preorder_left+size_left_subtree,inorder_left,inorder_root-1);root.right=myBuildTree(preorder,inorder,preorder_left+size_left_subtree+1,preorder_right,inorder_root+1,inorder_right);return root;}public TreeNode buildTree(int[] preorder, int[] inorder) {int n=inorder.length;indexMap=new HashMap<Integer,Integer>();for(int i=0;i<n;i++){indexMap.put(inorder[i],i);}return myBuildTree(preorder,inorder,0,n-1,0,n-1);}
}