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[leetcode] construct-binary-tree-from-preorder-and-inorder-
题目描述
Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: void ConstructTree(TreeNode *&tree, vector<int> &preorder, int prebegin, int preend, vector<int> &inorder, int inbegin, int inend) { if(prebegin > preend || inbegin > inend) return; int i; int mid = preorder[prebegin]; for(i = inbegin; i <= inend; i++) { if(inorder[i] == mid) break; } tree = new TreeNode(mid); ConstructTree(tree->left, preorder, prebegin+1, prebegin+i-inbegin, inorder, inbegin, i-1); ConstructTree(tree->right, preorder, prebegin+i+1-inbegin, preend, inorder, i+1, inend); } TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) { if(preorder.size() == 0 || inorder.size() == 0) return NULL; TreeNode *tree = NULL; ConstructTree(tree, preorder, 0, preorder.size()-1, inorder, 0, inorder.size()-1); return tree; } };
[leetcode] construct-binary-tree-from-preorder-and-inorder-
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