leetcode 105 Solution

代码解析

package com.demo.s105;

import java.util.HashMap;
import java.util.Map;

/**
 * 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。
 *
 * 来源：力扣（LeetCode）
 * 链接：https://leetcode.cn/problems/construct-binary-tree-from-preorder-and-inorder-traversal
 * 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
 */
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;
     }
 }

public class Solution {

    public TreeNode buildTree(int[] preorder, int[] inorder) {

        Map<Integer, Integer> cache = new HashMap<>();
        for(int i = 0; i< inorder.length;i++) {
            cache.put(inorder[i], i);
        }
        return buildTree(cache, preorder, inorder, 0, preorder.length, 0, inorder.length);

    }


    public TreeNode buildTree( Map<Integer, Integer> cache, int[] preorder, int[] inorder, int pres, int pree, int inos, int inoe) {
        if(pres >= pree) {
            return null;
        }
        int index = cache.get(preorder[pres]);
        TreeNode node = new TreeNode(preorder[pres]);
        node.left =  buildTree(cache, preorder, inorder, pres+1, pres+1 + index-inos, inos, index);
        node.right =  buildTree(cache, preorder, inorder, pres+1 + index-inos, pree,  index+1, inoe);
        return node;
    }
}