publicTreeNodebuildTree(int[] preorder,int[] inorder){if(preorder.length ==0)returnnull;introotIndex=0;inttempRootVal=preorder[rootIndex];inttempIndex=0;while(inorder[tempIndex] != tempRootVal) { tempIndex++; }TreeNoderoot=newTreeNode(tempRootVal);int[] newPre = Arrays.copyOfRange(pr...
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...
3的left - preOrder中3的下一个数字4, 范围缩小到inOrder position [0, -1],意味着走到底端了,返回到2;(3的right同理) 2的right - preOrder中3的下一个数字4, 范围缩小到inOrder position [2, 2],4在其中,设4为2的right,递归4; 4的left - preOrder中4的下一个数字5, 范围缩小到inOrder posit...
classSolution{public:TreeNode*buildTree(vector<int>&preorder,vector<int>&inorder){if(preorder.size()!=inorder.size()||preorder.size()==0||inorder.size()==0)returnNULL;else{returncreateTreeHelper(preorder,inorder,0,0,inorder.size()-1);}}TreeNode*createTreeHelper(vector<int>&preorder...
\ inorder:1 2 3 2 postorder: 3 2 1 \ 3 从上面我们可以看出,对于先序遍历都为 1 2 3 的五棵二叉树,它们的中序遍历都不相同,而它们的后序遍历却有相同的,所以只有和中序遍历一起才能唯一的确定一棵二叉树。但可能会有小伙伴指出,那第 889 题Construct Binary Tree from Preorder and Postorder Trav...
Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 题意;给定前序和中序遍历的序列,构造该二叉树 思路: 前序遍历的序列中每个元素都是一颗子树的根节点,从根节点开始到它的左子树的根节点... ...
https://oj.leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/ 解答: 树的问题一般都是递归解法或者BFS或者DFS。 注意到preorder的第一个节点肯定是root,然后在inorder里找root,就能确定左子树的长度。然后递归生成左子树和右子树。
buildSubTree(preorder,inorder,pre_start+leftlength+1,pre_end,pos+1,in_end);returnroot; } TreeNode* buildTree(vector<int>& preorder, vector<int>&inorder) {if(preorder.size()==0||inorder.size()==0)returnNULL;returnbuildSubTree(preorder,inorder,0,preorder.size()-1,0,inorder.size(...
public TreeNode buildTree(int[] preorder, int[] inorder) { int len =inorder.length; if(len==0) return null; TreeNode p = new TreeNode(preorder[0]); if(len==1) returnp; int index=0;for(;index<len;index++) {if(inorder[index]==preorder[0]) break;} ...
Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 题解: 利用preorder 和 inorder来构造树. e.g. preorder 1, 2, 4, 5, 3, 6, 7 inorder 4, 2, 5, 1, 6, 3, 7 ...