1.因为二叉搜索树的特性,将preorder数组排序,得到inorder。再将inorder的元素和下标用map存储起来,再对其进行递归。 2.利用二叉树的特性,初始化最小值,最大值,进行递归 3.用栈结构进行迭代。 classSolution {int[] preorder;intidx = 0; Map<Integer, Integer> map_inorder =newHashMap<>();publicTreeNode...
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...
intlen = preorder.length; TreeNode root = buildHelper(preorder,0, len -1, inorder,0, len -1); returnroot; } privateTreeNode buildHelper(int[] preorder,intpre_start,intpre_end, int[] inorder,intin_start,intin_end) { // bound if(pre_start > pre_end || in_start > in_end)...
* Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ classSolution{ public: voiddfs(vector<int>&preorder,TreeNode*root,intleft,intright){ if(left>right) ...
public TreeNode buildTree(int[] preorder, int[] inorder) { if(preorder==null || inorder==null) return null; HashMap<Integer, Integer> map = new HashMap<Integer, Integer>(); for(int i=0;i<inorder.length;i++) { map.put(inorder[i],i); ...
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...
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。
(preBegin == preEnd) return nullptr; //从中序遍历结果中找出根节点(根节点在前序遍历中是第一个节点) vector<int>::iterator itRoot = find(inBegin, inEnd, *preBegin); TreeNode *root = new TreeNode(*itRoot); //计算根的左子树节点个数 int leftSize = itRoot - inBegin; //在中序遍历...
# 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] :...