LeetCode-105-Construct a binary tree from pre-order and middle-order traversal sequence

Construct a binary tree from preorder and middle order traversal sequence

Title description: Given a tree, the pre-order traversal preorder and the middle-order traversal inorder . Please construct a binary tree and return its root node.

Please refer to LeetCode official website for example description.

Source: LeetCode
Link: https://leetcode-cn.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/
The copyright belongs to Lingkou Network. For commercial reprints, please contact the official authorization. For non-commercial reprints, please indicate the source.

Solution one: recursive method

To construct a binary tree by recursive method, the recursive process is as follows:

• If the pre-order traversal sequence or the middle-order traversal sequence is empty, return the empty tree directly;
• Because the first value of the preorder traversal sequence is the root node, the root node is obtained first;
• Then get the number of nodes leftCount and rightCount of the left and right subtrees of the root node according to the position of the root node in the in-order traversal;
• Then call this method recursively to get the left and right subtrees of the current root node;
• Finally, the root node is returned.

import com.kaesar.leetcode.TreeNode;

import java.util.Arrays;

public class LeetCode_105 {
    public static TreeNode buildTree(int[] preorder, int[] inorder) {
        // 当前序遍历序列或者中序遍历序列为空时，直接返回空树
        if (preorder == null || preorder.length == 0) {
            return null;
        }
        // 前序遍历序列的第一个值为根节点
        TreeNode root = new TreeNode(preorder[0]);
        // 左子树节点的数量
        int leftCount;
        // 中序遍历序列中，根节点左边的节点都是根节点左子树的节点
        for (leftCount = 0; leftCount < inorder.length; leftCount++) {
            if (inorder[leftCount] == preorder[0]) {
                break;
            }
        }
        // 根据左子树节点数和总的节点数计算右子树节点的数量
        int rightCount = inorder.length - 1 - leftCount;

        // 递归调用得到当前节点的左右子树
        root.left = buildTree(Arrays.copyOfRange(preorder, 1, leftCount + 1), Arrays.copyOfRange(inorder, 0, leftCount));
        root.right = buildTree(Arrays.copyOfRange(preorder, leftCount + 1, preorder.length), Arrays.copyOfRange(inorder, leftCount + 1, inorder.length));
        return root;
    }

    public static void main(String[] args) {
        int[] preorder = new int[]{3, 9, 20, 15, 7};
        int[] inorder = new int[]{9, 3, 15, 20, 7};
        buildTree(preorder, inorder).print();
    }
}

[Daily Message] My life has gone from hardship to ease without a strong heart. No one can accomplish anyone with one hand, the true God is in my heart. Only by working hard can we see the light!