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
publicclassBuildTreeFromPreorderAndInorder { publicTreeNode buildTree(int[] preorder,int[] inorder) { if(preorder ==null|| inorder ==null|| preorder.length ==0 || preorder.length != inorder.length) { returnnull; } intlen = preorder.length; TreeNode root = buildHelper(preorder,0,...
//测试 public class BuildTreeUsingInorderAndPreorder { public static void main(String[] args) { int[] preSort={1,2,4,7,3,5,6,8}; int[] inSort={4,7,2,1,5,3,8,6}; System.out.println(new Solution().buildTree(preSort, inSort)); } } //利用前序和中序重建二叉树 class Solutio...
inorder,(long)Integer.MAX_VALUE+1);}intpre=0;intin=0;privateTreeNodebuildTreeHelper(int[]preorder,int[]inorder,longstop){//到达末尾返回 nullif(pre==preorder.length){returnnull;}//到达停止点返回 null//当前停止点已经用了,in 后移if(inorder[in]==stop){in++;returnnull;}introot_val=pre...
preorder = [3,9,20,15,7] inorder = [9,3,15,20,7] 1. 2. Return the following binary tree: 3 / \ 9 20 / \ 15 7 1. 2. 3. 4. 5. 题目大意 使用先序遍历和中序遍历序列重构二叉树。 解题方法 递归 解决本题需要牢牢掌握先序遍历和中序遍历的含义,以及递归。
InTraverse(root); return 0; } */ Leecode AC代码: 代码语言:javascript 代码运行次数:0 运行 AI代码解释 classSolution{public:TreeNode*buildTree(vector<int>&preorder,vector<int>&inorder){if(preorder.size()!=inorder.size()||preorder.size()==0||inorder.size()==0)returnNULL;else{returncre...
current->right=build(p_l+left_tree_n+1,p_r,k+1,i_r); return current; } TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) { if(preorder.empty())return NULL; return build(preorder.begin(),preorder.end()-1,inorder.begin(),inorder.end()-1); ...
Is there a way to print out numbers for pre-order, post-order and in-order traversals of trees using just recursion and a number. The trees are binary and n is the numbe
preorder sequenceinorder sequenceIn m ist of the Data Structure textbooks there is a conclusion: "Given the preodcr and the inorder sequence of nodes, the binary tree structure can be determined exclusively".This paper gives a counter example to prove this conclusion is wrong.Authors also give...
preorder 的范围 [begin, end] 并不是 inorder 的范围,如果直接拿begin, end 来作为inorder[] 数组的边界条件,就会出错。所以必须要分为两个范围。 [preorderBegin, preorderEnd] [inorderBegin, inorderEnd] 这两个范围。 第二个错误,错的更深,其实到最后都没有彻底意识到,一定要当心,做此类题时。