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给定一个升序的数组,把他转换成一个高度平衡的二叉查找树
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Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
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递归的方法:
test.cpp:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 | #include <iostream> #include <cstdio> #include <stack> #include <vector> #include “BinaryTree.h” using namespace std; /** if(begin > end) TreeNode *sortedArrayToBST(vector< int> &num) vector<vector< int> > levelOrder(TreeNode *root) vector<vector< int> > matrix; vector<TreeNode *> path; int count = 1; if(count == 0) // 树中结点含有分叉, TreeNode *root = sortedArrayToBST(num); vector<vector< int> > ans = levelOrder(root); for ( int i = 0; i < ans.size(); ++i) |
结果输出:
4 2 6 1 3 5 7
BinaryTree.h:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 | #ifndef _BINARY_TREE_H_ #define _BINARY_TREE_H_ struct TreeNode TreeNode *CreateBinaryTreeNode( int value); #endif /*_BINARY_TREE_H_*/ |
BinaryTree.cpp:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 | #include <iostream> #include <cstdio> #include “BinaryTree.h” using namespace std; /** //创建结点 return pNode; //连接结点 //打印节点内容以及左右子结点内容 if(pNode->left != NULL) if(pNode->right != NULL) printf( “\n”); //前序遍历递归方法打印结点内容 if(pRoot != NULL) if(pRoot->right != NULL) void DestroyTree(TreeNode *pRoot) delete pRoot; DestroyTree(pLeft); |