题解 | #重建二叉树#

重建二叉树

https://www.nowcoder.com/practice/8a19cbe657394eeaac2f6ea9b0f6fcf6

import java.util.*;

/*
 * public class TreeNode {
 *   int val = 0;
 *   TreeNode left = null;
 *   TreeNode right = null;
 *   public TreeNode(int val) {
 *     this.val = val;
 *   }
 * }
 */

public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可
     *
     *
     * @param preOrder int整型一维数组
     * @param vinOrder int整型一维数组
     * @return TreeNode类
     */
    public TreeNode reConstructBinaryTree (int[] preOrder, int[] vinOrder) {
        // write code here
        if (preOrder.length == 0 || vinOrder.length == 0) {
            return null;
        }
        return buildTree(preOrder, 0, preOrder.length - 1, vinOrder, 0, vinOrder.length - 1);

    }

    public TreeNode buildTree(int[] preOrder, int prestart, int preend,
                              int[] vinOrder, int vinstart, int vinend) {
        if (prestart > preend || vinstart > vinend) {
            return null;
        }
        TreeNode head = new TreeNode(preOrder[prestart]);
        int rootindex = vinstart;
        while(rootindex <= vinend && vinOrder[rootindex] != preOrder[prestart]) {
            rootindex++;
        }

        head.left = buildTree(preOrder, prestart + 1, prestart + rootindex - vinstart, vinOrder, vinstart, rootindex - 1);
        head.right = buildTree(preOrder, prestart + rootindex - vinstart + 1, preend, vinOrder, rootindex + 1, vinend);
        return head;
    }
}

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