[LeetCode]110. Balanced Binary Tree

[LeetCode]110. Balanced Binary Tree

题目描述

思路

代码

#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>

using namespace std;

struct TreeNode {
int val;
TreeNode* left;
TreeNode* right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

class Solution {
public:
TreeNode* vectorToTree(vector<int> nums){
if (nums.empty())
return NULL;
TreeNode* root = new TreeNode(nums[0]);
queue<TreeNode*> q;
q.push(root);

for (int i = 1; i < nums.size();) {
int curSize = q.size();
while (curSize--) {
TreeNode* node = q.front();
q.pop();

if (nums[i]) {
node->left = new TreeNode(nums[i]);
q.push(node->left);
}
else
node->left = NULL;
++i;
if (i == nums.size())
break;
if (nums[i]) {
node->right = new TreeNode(nums[i]);
q.push(node->right);
}
else
node->right = NULL;
++i;
if (i == nums.size())
break;
}
}
return root;
}

bool isBalanced(TreeNode* root) {
maxDepth(root);
return flag;
}
int maxDepth(TreeNode* root) {
if (root == NULL)
return 0;
int leftDepth = maxDepth(root->left),
rightDepth = maxDepth(root->right);
if (abs(leftDepth - rightDepth) > 1)
flag = false;
return max(leftDepth + 1, rightDepth + 1);
}
private:
bool flag = true;
};

int main() {
vector<int> nums = { 1, 0, 1, 1 };
Solution s;
TreeNode* root = s.vectorToTree(nums);
cout << s.isBalanced(root) << endl;

system("pause");
return 0;
}