65publicstaticvoidinOrder(TreeNode root){66if(root ==null)return;67inOrder(root.left);68visit(root);69inOrder(root.right);70}7172publicstaticvoidinOrder2(TreeNode root){73if(root ==null)return;74Stack<TreeNode> stack =newStack<TreeNode>();75while(!stack.empty() || root !=null){76...
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<...
* Val int * Left *TreeNode * Right *TreeNode * } */funcbuildTree(preorder[]int,inorder[]int)*TreeNode{inPos:=make(map[int]int)fori:=0;i<len(inorder);i++{inPos[inorder[i]]=i}returnbuildPreIn2TreeDFS(preorder,0,len(preorder)-1,0,inPos)}funcbuildPreIn2TreeDFS(pre[]int,...
*/classSolution{public TreeNodebuildTree(int[]preorder,int[]inorder){Map<Integer,Integer>inorderMap=newHashMap<Integer,Integer>();for(int i=0;i<inorder.length;i++)inorderMap.put(inorder[i],i);returnbuildTree(preorder,0,preorder.length-1,inorder,0,inorder.length-1,inorderMap);}publ...
[preorderBegin]){rootIndex=i;break;}}int leftTreeSize=rootIndex-inorderBegin;root.left=buildTree(preorderBegin+1,preorderBegin+leftTreeSize,inorderBegin,rootIndex-1,preorder,inorder);root.right=buildTree(preorderBegin+leftTreeSize+1,preorderEnd,rootIndex+1,inorderEnd,preorder,inorder);...
Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. recursive answer: 递归方法:在preorder找到root节点,接着在inorder里找到root节点,则inorder被分为两部分,一部分是left,一部分是right。在确定right支的时候要注...
1.Given preorder and inorder traversal of a tree, construct the binary tree. 给定一个二叉树的前序和中序遍历,重建这棵二叉树。 2.Enter the tree traversal Preorder traversal of the results and the results of established tree and gives the results after the traversal. 输入树的前序遍历结果...
INORDER AND PREORDER TRAVERSAL and DISPLAY OF EXISTING BINARY TREE IN COMPUTER MEMORYP K KumaresanJournal of Harmonized Research
就是倒序输出了数组的元素，static int t刚开始下标是3，y = s[t--],t变成2，call返回s[3]的值，下一次进入call,t的值就是2，然后t变成1，然后返回s[2],依次类推
T=( Bitree)ma11oc{ sizeof( Bi Tnode));//申请结点空间 root=find (Inorder iinb. ine], Preorder [preb]); //忽略了实现细节,找到根结点在中序中的位置 T->data=Preorder [pre] if (preb libegin=inb//左子树中序序列的起始位置