public class BuildTreeFromPreorderAndInorder { public TreeNode buildTree(int[] preorder, int[] inorder) { if (preorder == null || inorder == null || preorder.length == 0 || preorder.length != inorder.length) { return null; ...
TreeNode*buildTree(vector<int> &preorder,intpLeft,intpRight, vector<int> &inorder,intiLeft,intiRight) {if(pLeft > pRight || iLeft > iRight)returnNULL;inti =0;for(i = iLeft; i <= iRight; ++i) {if(preorder[pLeft] == inorder[i])break; } TreeNode*cur =newTreeNode(preorder...
for(int i=0;i<inorder.length;i++) { map.put(inorder[i],i); } return helper(preorder,0,preorder.length-1,inorder,0,inorder.length-1, map); } private TreeNode helper(int[] preorder, int preL, int preR, int[] inorder, int inL, int inR, HashMap<Integer, Integer> map) {...
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...
intsearchNode(intinorder[],intinorderSize,intkey){inti;for(i=0;i<inorderSize;i++){if(key==inorder[i]){returni;}}return-1;}structTreeNode*buildTree(int*preorder,intpreorderSize,int*inorder,intinorderSize){if(preorder==NULL||inorder==NULL||preorderSize==0||inorderSize==0)return...
* TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */publicclassSolution{public TreeNodebuildTree(int[]preorder,int[]inorder){if(preorder==null||preorder.length==0)returnnull;if(inorder==null||inorder.length==0)returnnull;if(preorder.length!=inorder.length...
Construct Binary Tree from Preorder and Inorder Traversal 题目描述(中等难度) 根据二叉树的先序遍历和中序遍历还原二叉树。 解法一 递归 先序遍历的顺序是根节点,左子树,右子树。中序遍历的顺序是左子树,根节点,右子树。 所以我们只需要根据先序遍历得到根节点,然后在中序遍历中找到根节点的位置,它的左边就...
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...
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。
# 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] :...