inorder,(long)Integer.MAX_VALUE+1);}intpre=0;intin=0;privateTreeNodebuildTreeHelper(int[]preorder,int[]inorder,longstop){//到达末尾返回 nullif(pre==preorder.length){returnnull;}//到达停止点返回 null//当前停止点已经用了,in 后移if(inorder[in]==stop){in++;returnnull;}introot_val=pre...
left(NULL), right(NULL) {}8* };9*/10classSolution {11public:12TreeNode* buildTree(vector<int>& preorder, vector<int>&inorder) {13returnbuild(preorder, inorder,0, preorder.size() -1,0, inorder.size() -1);14
Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. ++++++++++++++++++++++++++++++++++++++++ test.cpp: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25...
preorder边界是来确定左子树的左右子树边界,后面的inorder边界来获取左子树大小,知道左子树大小也知道了右子树大小。 root->left = bT(preorder,inorder,preorder_left+1,preorder_left+left_size,inorder_left,inorder_root-1); //preorder_left+left_size+1---preorder_right是preorder的右子树位置,对应in...
public TreeNode buildTree(int[] preorder, int[] inorder) { if(preorder==null || inorder==null) return null; HashMap<Integer, Integer> map = new HashMap<Integer, Integer>(); for(int i=0;i<inorder.length;i++) { map.put(inorder[i],i); ...
preoder保存的都是头结点,依次取出来作为ROOT节点,然后在inorder数组里面找出当前preoder[i]的下标,之后在[inorder+index+1, end]...
preorder 的范围 [begin, end] 并不是 inorder 的范围,如果直接拿begin, end 来作为inorder[] 数组的边界条件,就会出错。所以必须要分为两个范围。 [preorderBegin, preorderEnd] [inorderBegin, inorderEnd] 这两个范围。 第二个错误,错的更深,其实到最后都没有彻底意识到,一定要当心,做此类题时。
Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume thatduplicates do not existin the tree. 思路 主要是根据前序遍历和中序遍历的特点解决这个题目。 1、确定树的根节点。树根是当前树中所有元素在前序遍历中最先出现的元素。
https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/description/ """ # Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution(object): d...
Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 做了第106题再来做这个就简单了,写递归的话思路一模一样: /** * Definition for a binary tree node. ...