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Construct Binary Tree from Preorder and Inorder Traversal<leetcode>
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:知道二叉树的前序遍历和中序遍历,重新构建出二叉树。
算法:(略),代码如下:
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */10 class Solution {11 public:12 TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {13 if(preorder.empty()||inorder.empty()) return NULL;14 return build(preorder,0,preorder.size()-1,inorder,0,inorder.size()-1);15 }16 TreeNode *build(vector<int> &preorder,int l,int r,vector<int> &inorder,int ll,int rr)17 {18 if(l>r||ll>rr) return NULL;19 TreeNode* root=new TreeNode(0);20 root->val=preorder[l];21 if(l==r) return root;22 int mid=0;23 for(int i=ll;i<=rr;i++)24 {25 if(inorder[i]==preorder[l])26 {27 mid=i;28 break;29 }30 }31 int midd=mid-ll+l;32 root->left=build(preorder,l+1,midd,inorder,ll,mid-1);33 root->right=build(preorder,midd+1,r,inorder,mid+1,rr);34 return root;35 }36 };
Construct Binary Tree from Preorder and Inorder Traversal<leetcode>
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