Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
#include <iostream> #include <algorithm> #include <vector> /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ using std::vector; using std::find; struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int x) : val(x), left(NULL), right(NULL) {} }; class Solution { private: TreeNode* buildTree(vector<int>::iterator PreBegin, vector<int>::iterator PreEnd, vector<int>::iterator InBegin, vector<int>::iterator InEnd) { if (PreBegin == PreEnd) { return NULL; } int HeadValue = *PreBegin; TreeNode *HeadNode = new TreeNode(HeadValue); vector<int>::iterator LeftEnd = find(InBegin, InEnd, HeadValue); if (LeftEnd != InEnd) { HeadNode->left = buildTree(PreBegin + 1, PreBegin + (LeftEnd - InBegin) + 1, InBegin, LeftEnd); } HeadNode->right = buildTree(PreBegin + (LeftEnd - InBegin) + 1, PreEnd, LeftEnd + 1, InEnd); return HeadNode; } public: TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) { if (preorder.empty()) { return NULL; } return buildTree(preorder.begin(), preorder.end(), inorder.begin(), inorder.end()); } };
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LeetCode_Construct Binary Tree from Preorder and Inorder Traversal
原文地址:http://blog.csdn.net/sheng_ai/article/details/46684535