标签：div leetcode hat back col duplicate sum uil tco
Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
For example, given
preorder = [3,9,20,15,7] inorder = [9,3,15,20,7]
Return the following binary tree:
3
/ 9 20
/ 15 7
已知二叉树的前序遍历这中序遍历，求二叉树，这个其实挺简单的，在前序遍历中取第一个元素，然后在中序遍历找到相应的元素，左边的为左子树，右边的为右子树，根据长度在前序遍历中找到相应左子树和右子树的前序遍历。然后就可以递归了。
class Solution { public: TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) { vector<int> leftpre; vector<int> leftino; vector<int> rightpre; vector<int> rightino; if(preorder.empty()||inorder.empty()) return NULL; int root = preorder[0]; int left=0;int right=0;int tag=0; for(int i=0;i<inorder.size();i++) { if(inorder[i]==root) { tag=1; } else if(tag==0) {left++;leftino.push_back(inorder[i]);} else {right++;rightino.push_back(inorder[i]);} } for(int i=1;i<left+1;i++) { leftpre.push_back(preorder[i]); } for(int i=left+1;i<preorder.size();i++) { rightpre.push_back(preorder[i]); } TreeNode* tree = new TreeNode(root); tree->left = buildTree(leftpre,leftino); tree->right = buildTree(rightpre,rightino); return tree; } };
LeetCode 105. Construct Binary Tree from Preorder and Inorder Traversal
标签：div leetcode hat back col duplicate sum uil tco
原文地址：https://www.cnblogs.com/dacc123/p/9215539.html
