力扣第105题-从前序与中序遍历序列构造二叉树

题目
链接

题解

import java.util.Map;
import java.util.HashMap;
/**
 * 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) {
        // 哈希表，k：中序遍历的节点，v：中序遍历的下标
        Map<Integer, Integer> hash = new HashMap<>(inorder.length);
        for (int i = 0; i < inorder.length; i++) {
            hash.put(inorder[i], i);
        }
        return build(preorder, inorder, 0, 0, inorder.length - 1, hash);
    }

    private TreeNode build(int[] preorder, int[] inorder, int rootPreorderIndex, int inBegin, int inEnd, Map<Integer, Integer> hash) {
        if (inBegin == inEnd) {
            return new TreeNode(inorder[inBegin], null, null);
        }
        int rootValue = preorder[rootPreorderIndex];
        // 找根节点在中序遍历中的下标
        int rootInorderIndex = hash.get(rootValue);
        // 递归左子树
        TreeNode left = null;
        if (inBegin <= rootInorderIndex - 1 && rootPreorderIndex + 1 <= preorder.length - 1) {
            left = build(preorder, inorder, rootPreorderIndex + 1, inBegin, rootInorderIndex - 1, hash);
        }
        // 递归右子树
        TreeNode right = null;
        if (rootInorderIndex + 1 <= inEnd) {
            // 计算右子树根节点在前序遍历中的下标，计算公式：当前根节点在前序遍历中的下标 + 左子树节点个数
            // 左子树节点个数计算公式：当前根节点在中序遍历中的下标 - 中序遍历起始下标 + 1
            int rightRootPreorderIndex = rootPreorderIndex + (rootInorderIndex - inBegin) + 1;
            right = build(preorder, inorder, rightRootPreorderIndex, rootInorderIndex + 1, inEnd, hash);
        }
        return new TreeNode(preorder[rootPreorderIndex], left, right);
    }

    private int seach(int[] src, int target, int start, int end) {
        for (int i = start; i <= end; i++) {
            if (src[i] == target) {
                return i;
            }
        }
        return -1;
    }

}