105. 从前序与中序遍历序列构造二叉树

Posted 2021-11-02 潜行前行

105. 从前序与中序遍历序列构造二叉树

给定一棵树的前序遍历 preorder 与中序遍历 inorder。请构造二叉树并返回其根节点。

示例 1:

Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
Output: [3,9,20,null,null,15,7]
示例 2:

Input: preorder = [-1], inorder = [-1]
Output: [-1]

/**
 * 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 {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if(preorder.length == 0)
            return null;
        if(preorder[0] == inorder[0]){
            TreeNode root = new TreeNode(preorder[0]);
            if(preorder.length > 1){
                TreeNode right = buildTree(Arrays.copyOfRange(preorder,1,preorder.length),Arrays.copyOfRange(inorder,1,preorder.length));
                root.right = right; 
                return root;
            }
        }
        for( int i=0;i<preorder.length;i++){
            if(preorder[0] == inorder[i]){
                TreeNode left = buildTree(Arrays.copyOfRange(preorder,1,i+1),Arrays.copyOfRange(inorder,0,i));
                TreeNode right = buildTree(Arrays.copyOfRange(preorder,i+1,preorder.length),Arrays.copyOfRange(inorder,i+1,preorder.length));
                TreeNode root = new TreeNode(preorder[0]);
                root.left = left;
                root.right = right;
                return root;
            }
        }
        return null;
    }
}