《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：

 // 4.1 Implement an algorithm to check if a bianry tree is height-balanced.
 #include <algorithm>
 #include <cstdio>
 #include <unordered_map>
 using namespace std;

 struct TreeNode {
     int val;
     TreeNode *left;
     TreeNode *right;

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

 void constructBinaryTree(TreeNode *&root)
 {
     int val;

     scanf("%d", &val);
     if (val <= ) {
         root = nullptr;
     } else {
         root = new TreeNode(val);

         constructBinaryTree(root->left);
         constructBinaryTree(root->right);
     }
 }

 void clearBinaryTree(TreeNode *&root) {
     if (root == nullptr) {
         return;
     } else {
         clearBinaryTree(root->left);
         clearBinaryTree(root->right);
         delete root;
         root = nullptr;
     }
 }

 void calcHeights(TreeNode *root, unordered_map<TreeNode *, int> &heights)
 {
     if (root == nullptr) {
         heights[root] = ;
     } else {
         calcHeights(root->left, heights);
         calcHeights(root->right, heights);
         heights[root] = max(heights[root->left], heights[root->right]) + ;
     }
 }

 bool isBalanced(TreeNode *root, unordered_map<TreeNode *, int> &heights)
 {
     if (root == nullptr) {
         return true;
     } else {
         return abs(heights[root->left] - heights[root->right]) <= ;
     }
 }

 int main()
 {
     TreeNode *root;
     unordered_map<TreeNode *, int> heights;

     while (true) {
         constructBinaryTree(root);
         if (root == nullptr) {
             break;
         }

         calcHeights(root, heights);
         if (isBalanced(root, heights)) {
             printf("Yes\n");
         } else {
             printf("No\n");
         }
         heights.clear();
         clearBinaryTree(root);
     }

     return ;
 }