0,preorder.length,inorder,0,inorder.length);}privateTreeNodebuildTreeHelper(int[]preorder,intp_start,intp_end,int[]inorder,inti_start,inti_end){// preorder 为空,直接返回 nullif(p_start==p_end){returnnull;}introot_val=preorder[p_start];TreeNoderoot=newTreeNode(root_val);//在中序遍...
root.right= self.buildTree(preorder[i+1:], inorder[i+1:])returnroot 同样的方法做106. Construct Binary Tree from Inorder and Postorder Traversal,后序和中序,建二叉树,根节点是后序的最后一个元素。 classSolution(object):defbuildTree(self, inorder, postorder):""":type inorder: List[int] ...
TreeNode* build(vector<int> &preorder,vector<int> &inorder){if(preorder.size() ==0|| inorder.size() ==0)returnNULL;//递归条件。当然,也还可以加上if(preorder.size() == 1 || inorder.size() == 1) return cur;这个递归条件。introotval = preorder[0];inti =0;for(i =0; i <...
TreeNode* buildTree(vector<int>& preorder, int preBeg, vector<int>& inorder, int inBeg, int size) { if(size <= 0) return nullptr; TreeNode* curNode = new TreeNode(preorder[preBeg]); int inorderLeftTreeEnd = inBeg; while(inorder[inorderLeftTreeEnd] != preorder[preBeg]) ++in...
Given preorder and inorder traversal of a tree, construct the binary tree. 本题就是根据前序遍历和中序遍历的结果还原一个二叉树。 题意很简答,就是递归实现,直接参考代码吧。 查询index部分可以使用Map做查询,建议和下一道题 leetcode 106. Construct Binary Tree from Inorder and Postorder Traversal 中...
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; ...
TreeNode root=newTreeNode(preorder[preorderBegin]);int rootIndex=0;for(int i=inorderBegin;i<=inorderEnd;i++){if(inorder[i]==preorder[preorderBegin]){rootIndex=i;break;}}int leftTreeSize=rootIndex-inorderBegin;root.left=buildTree(preorderBegin+1,preorderBegin+leftTreeSize,inorderBegin...
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...
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。
# 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): def buildTree(self, preorder, inorder): """ :type preorder: List[int] :type inorder: List[int] :...