leetCode 105.Construct Binary Tree from Preorder and Inorder Traversal (根據前序遍歷和中序遍歷構造二叉樹)

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

Note:

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

思路：首先根據前序遍歷得到根節點，然後在中序遍歷中得到根節點的位置，左邊的爲左子樹，右邊的爲右子樹。

然後再遞歸求解左子樹和右子樹的構造即可。代碼如下：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        /**
         * 1.根據前序遍歷，先確定根節點
         * 2.然後在中序遍歷中查找根節點，確定根節點在中序遍歷的位置
         * 3.根據索引位置分割左右子樹的前序和中序遍歷
         * 4.遞歸求解根節點的左右子樹
         */ 
        if(preorder.length == 0 || inorder.length == 0)
            return null;
        TreeNode root = new TreeNode(preorder[0]);
        int k = 0;
        for(; k < inorder.length; k++){
            if(inorder[k] == preorder[0]){
                break;
            }
        }
        int[] p1 = Arrays.copyOfRange(preorder,1,k+1);
        int[] q1 = Arrays.copyOfRange(preorder,k+1,preorder.length);
        int[] p2 = Arrays.copyOfRange(inorder,0,k);
        int[] q2 = Arrays.copyOfRange(inorder,k+1,inorder.length);
        
        root.left = buildTree(p1,p2);
        root.right = buildTree(q1,q2);
        return root;
    }
}