LeetCode 105. 从前序与中序遍历序列构造二叉树

105. 从前序与中序遍历序列构造二叉树

给定两个整数数组?preorder?和?inorder?，其中?preorder?是二叉树的先序遍历，?inorder?是同一棵树的中序遍历，请构造二叉树并返回其根节点。

示例 1:

输入: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
输出: [3,9,20,null,null,15,7]

示例 2:

输入: preorder = [-1], inorder = [-1]
输出: [-1]

提示:

• 1 <= preorder.length <= 3000
• inorder.length == preorder.length
• -3000 <= preorder[i], inorder[i] <= 3000
• preorder?和?inorder?均?无重复?元素
• inorder?均出现在?preorder
• preorder?保证?为二叉树的前序遍历序列
• inorder?保证?为二叉树的中序遍历序列

解法思路：

递归

?

/**
 * 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 {
    // Recursion
    // Time: O(n)
    // Space: O(n)
    private Map<Integer, Integer> indexMap;

    public TreeNode buildTree(int[] preorder, int[] inorder) {
        // 使用 Map 存储中序遍历数组中每个节点值对应的索引，以便快速查找
        indexMap = new HashMap<>();
        for (int i = 0; i < inorder.length; i++) {
            indexMap.put(inorder[i], i);
        }
        return buildTreeHelper(preorder, inorder, 0, 0, inorder.length - 1);
    }

    private TreeNode buildTreeHelper(int[] preorder, int[] inorder, int preStart, int inStart, int inEnd) {
        // 递归终止条件：先序遍历数组为空或中序遍历数组的范围越界
        if (preStart > preorder.length - 1 || inStart > inEnd) {
            return null;
        }

        // 当前子树的根节点值为先序遍历数组的第一个元素
        int rootVal = preorder[preStart];
        TreeNode root = new TreeNode(rootVal);

        // 在中序遍历数组中找到根节点值的索引
        int inIndex = indexMap.get(rootVal);

        // 递归构建左右子树
        root.left = buildTreeHelper(preorder, inorder, preStart + 1, inStart, inIndex - 1);
        root.right = buildTreeHelper(preorder, inorder, preStart + inIndex - inStart + 1, inIndex + 1, inEnd);

        return root;
    }
}

?