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

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

简单的思想：正序排列时，某节点的右节点一定是该节点的右子树，中序排列，某节点的左节点一定是该节点的左子树；

深搜递归

/**
 * 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[] preorder, int[] inorder) {
	    return buildTree(preorder,inorder,0,preorder.length-1,0,inorder.length-1);
	}
	
	private TreeNode buildTree(int []preorder, int []inorder,int pst,int pend,int inst, int inend){
		if(pst>pend||inst>inend||preorder.length<1){
			return null;
		}
		TreeNode tn =new TreeNode(preorder[pst]);
		int index = -1;
		for(int i=inst;i<=inend;i++){
			if(inorder[i] == preorder[pst]){
				index = i;
				break;
			}
		}
		tn.left = buildTree(preorder,inorder,pst+1,pst+index-inst,inst,inst+index-1);
		tn.right = buildTree(preorder,inorder,pst+index-inst+1,pend,index+1,inend);
		return tn;
	}
}

[LeetCode]Construct Binary Tree from Preorder and Inorder Traversal