DFS练习

105 从前序与中序遍历序列构造二叉树

在这里插入图片描述

import java.util.HashMap;
import java.util.Map;class TreeNode {int val;TreeNode left;TreeNode right;public TreeNode(int val) {this.val = val;}
}public class Letcode105 {public TreeNode bulidTree(int[] preOrder, int[] inOrder) {int preLength = preOrder.length;int inLength = inOrder.length;// 用map存储中序遍历的数组及对应的下表Map<Integer, Integer> map = new HashMap<>();for (int i = 0; i < inLength; i++) {map.put(inOrder[i], i);}return bulidTree(preOrder, 0, preLength - 1,map, 0, inLength - 1);}private TreeNode bulidTree(int[] preOrder, int preLeft, int preRight, Map<Integer, Integer> map, int inLeft, int inRight) {// 不管是什么序，左边的边界一定小于右边的边界if (preLeft > preRight || inLeft > inRight) {return null;}// 前序遍历，第一个左边界一定是根节点int root = preOrder[preLeft];// 新建树TreeNode treeNode = new TreeNode(root);int pIndex = map.get(root);treeNode.left = bulidTree(preOrder, preLeft + 1, pIndex - inLeft + preLeft, map,inLeft, pIndex - 1);treeNode.right = bulidTree(preOrder,pIndex - inLeft + preLeft + 1, preRight,map,pIndex + 1,inRight);return treeNode;}
}