1/**2* Definition for a binary tree node.3* public class TreeNode {4* int val;5* TreeNode left;6* TreeNode right;7* TreeNode(int x) { val = x; }8* }9*/10classSolution11{12publicTreeNode buildTree(int[] preorder,int[] inorder)13{14Map<Integer, Integer> inMap =newHashMap<...
publicTreeNode buildTree(int[] preorder,int[] inorder) { returnhelper(0,0, inorder.length -1, preorder, inorder); } publicTreeNode helper(intpreStart,intinStart,intinEnd,int[] preorder,int[] inorder) { if(preStart > preorder.length -1|| inStart > inEnd) { returnnull; } Tree...
TreeNode getRootByRecursion(int[] preorder, int[] inorder, int preStart, int inStart, int inEnd) { if(inStart > inEnd) return null; else { //前序的第一个节点就是父节点 TreeNode root=new TreeNode(preorder[preStart]); int index=getIndex(inorder,preorder[preStart]); //lenLeft是...
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...
Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 思路 这道题目,翻译一下就是:如何根据先序和中序遍历的数组,来构造二叉树。其中Note部分说,假设不存在重复项。为什么这样说呢?因为如果存在重复项,那么构造出来...
Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 思想: 就按照中序遍历和后序遍历建立二叉树 C++代码: /** * Definition for binary tree * struct TreeNode { ...
* 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...
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. ...
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。