Given preorder and inorder traversal of a tree, construct the binary tree.
这道题是要根据先序遍历和中序遍历构造出一棵二叉树，这道题和上一题是一样的。上一题不过是通过后序遍历和中序遍历构造一棵二叉树。
只要将代码稍稍修改即可
1 public class Solution { 2 public TreeNode buildTree(int[] preorder, int[] inorder) { 3 TreeNode root; 4 if(preorder.length == 0) 5 return null; 6 return createTree(preorder, inorder, 0, preorder.length - 1, 0, inorder.length - 1); 7 8 } 9 private TreeNode createTree(int[] preorder, int[] inorder, int preStart, int preEnd, int inStart ,int inEnd){ 10 if(inStart > inEnd) 11 return null; 12 int value_root = preorder[preStart]; //根据先序遍历第一个元素为根节点,找出根节点的值 13 int index = 0; //根节点在中序遍历中的索引值 14 for(int i = inStart; i <= inEnd; i++){ //查找根节点在中序遍历中的索引值 15 if(value_root == inorder[i]){ 16 index = i; 17 break; 18 } 19 }//for 20 int length_leftTree = index - inStart; //左子树的长度 21 TreeNode root_left = createTree(preorder, inorder, preStart + 1, preStart + length_leftTree, inStart, index - 1); 22 TreeNode root_right = createTree(preorder, inorder, preStart + length_leftTree + 1, preEnd, index + 1, inEnd); 23 24 TreeNode root = new TreeNode(value_root); 25 root.left = root_left; 26 root.right = root_right; 27 28 return root; 29 } 30 }