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...
}privateTreeNode buildTree(int[] preorder,intpreLo,intpreHi,int[] inorder,intinLo,intinHi) {if(preLo > preHi || inLo >inHi)returnnull; TreeNode root=newTreeNode(preorder[preLo]);introotAtInorder = 0;for(inti = inLo; i <= inHi; i++) {if(inorder[i] == root.val) {//...
Given preorder and inorder traversal of a tree, construct the binary tree. 解决思路 首先确定根节点,然后确定左右子树的节点数目。依次递归即可。 假设输入的序列均合法。 程序 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 ...
vector<int>& inorder, int inBeg, int size) { if(size <= 0) return nullptr; TreeNode* curNode = new TreeNode(preorder[preBeg]); int inorderLeftTreeEnd = inBeg; while(inorder[inorderLeftTreeEnd] != preorder[preBeg]) ++inorderLeftTreeEnd; int leftTreeSize = inorderLeftTreeEnd -...
105. Construct Binary Tree from Preorder and Inorder Traversal,题目Givenpreorderandinordertraversalofatree,constructthebinarytree.Note:Youmayassumethatduplicatesdonotexistinthet
题目:105. Construct Binary Tree from Preorder and Inorder Traversal Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 翻译:给定树的前序和中序遍历,构造二叉树。 注意: 树中不存在重复项。
* Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */publicclassSolution{public TreeNodebuildTree(int[]preorder,int[]inorder){if(preorder==null||preorder.length==0)returnnull;if(in...
* 题目: 105.Construct Binary Tree from Preorder and Inorder Traversal * 网址:https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/ * 结果:AC * 来源:LeetCode * 博客: ---*/#include<iostream>#include<vector>#include<algorithm>usingnamespacestd;structTree...
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...
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。