TreeNode(intx) { val = x; } } publicstaticTreeNode buildTree(int[] preorder,int[] inorder) { if(preorder==null||inorder==null||preorder.length!=inorder.length) returnnull; returnbuild(preorder, inorder,0, preor
题目: Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 根据前序遍历和中序遍历结果构造二叉树。 思路分析: 分析二叉树前序遍历和中序遍历的结果我们发现: 二叉树前序遍历的第一个节点是根节点。 在中序遍历...
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...
publicTreeNode buildTree(int[] preorder,int[] inorder) { if(preorder ==null|| inorder ==null|| preorder.length ==0 || preorder.length != inorder.length) { returnnull; } intlen = preorder.length; TreeNode root = buildHelper(preorder,0, len -1, inorder,0, len -1); returnr...
105. Construct Binary Tree from Preorder and Inorder Traversal,思路:对于一棵树,可以看成许多小树组成,每棵小树都有自己的root,我们从这里入手。对于每棵小树我们都需要定位其root,对于preorder,第一个元素就是root,但inorder还需要查找,但如果每次都遍历搜索
105. Construct Binary Tree from Preorder and Inorder Traversal——tree,程序员大本营,技术文章内容聚合第一站。
private TreeNode helper(int[] preorder, int preL, int preR, int[] inorder, int inL, int inR, HashMap<Integer, Integer> map) { if(preL>preR || inL>inR) return null; TreeNode root = new TreeNode(preorder[preL]); int index = map.get(root.val); ...
InTraverse(root); return 0; } */ Leecode AC代码: 代码语言:javascript 代码运行次数:0 运行 AI代码解释 classSolution{public:TreeNode*buildTree(vector<int>&preorder,vector<int>&inorder){if(preorder.size()!=inorder.size()||preorder.size()==0||inorder.size()==0)returnNULL;else{returncre...
Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 翻译:给定树的前序和中序遍历,构造二叉树。 注意: 树中不存在重复项。 思路:首先,你应该知道 前序遍历:根节点,左子树,右子树; ...
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...