《Cracking the Coding Interview》——第4章：树和图——题目1

2014-03-19 03:30

题目：判断一个二叉树是否为平衡二叉树，即左右子树高度相差不超过1。

解法：递归算高度并判断即可。

代码：

 1 // 4.1 Implement an algorithm to check if a bianry tree is height-balanced.

 2 #include <algorithm>

 3 #include <cstdio>

 4 #include <unordered_map>

 5 using namespace std;

 6 

 7 struct TreeNode {

 8     int val;

 9     TreeNode *left;

10     TreeNode *right;

11     

12     TreeNode(int _val = 0): val(_val), left(nullptr), right(nullptr) {};

13 };

14 

15 void constructBinaryTree(TreeNode *&root)

16 {

17     int val;

18     

19     scanf("%d", &val);

20     if (val <= 0) {

21         root = nullptr;

22     } else {

23         root = new TreeNode(val);

24 

25         constructBinaryTree(root->left);

26         constructBinaryTree(root->right);

27     }

28 }

29 

30 void clearBinaryTree(TreeNode *&root) {

31     if (root == nullptr) {

32         return;

33     } else {

34         clearBinaryTree(root->left);

35         clearBinaryTree(root->right);

36         delete root;

37         root = nullptr;

38     }

39 }

40 

41 void calcHeights(TreeNode *root, unordered_map<TreeNode *, int> &heights)

42 {

43     if (root == nullptr) {

44         heights[root] = 0;

45     } else {

46         calcHeights(root->left, heights);

47         calcHeights(root->right, heights);

48         heights[root] = max(heights[root->left], heights[root->right]) + 1;

49     }

50 }

51 

52 bool isBalanced(TreeNode *root, unordered_map<TreeNode *, int> &heights)

53 {

54     if (root == nullptr) {

55         return true;

56     } else {

57         return abs(heights[root->left] - heights[root->right]) <= 1;

58     }

59 }

60 

61 int main()

62 {

63     TreeNode *root;

64     unordered_map<TreeNode *, int> heights;

65     

66     while (true) {

67         constructBinaryTree(root);

68         if (root == nullptr) {

69             break;

70         }

71         

72         calcHeights(root, heights);

73         if (isBalanced(root, heights)) {

74             printf("Yes\n");

75         } else {

76             printf("No\n");

77         }

78         heights.clear();

79         clearBinaryTree(root);

80     }

81     

82     return 0;

83 }