110. Balanced Binary Tree - easy
문제
Given a binary tree, determine if it is height-balanced.
For this problem, a height-balanced binary tree is defined as:
a binary tree in which the left and right subtrees of every node differ in height by no more than 1.
제한사항
입출력 예
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Example 1:
Given the following tree [3,9,20,null,null,15,7]:
3
/ \
9 20
/ \
15 7
Return true.
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Example 2:
Given the following tree [1,2,2,3,3,null,null,4,4]:
1
/ \
2 2
/ \
3 3
/ \
4 4
Return false.
풀이
- Tree, DFS
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/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int dfs(TreeNode* root) {
if (!root) {
return 0;
}
int depth = 0;
std::stack<std::pair<TreeNode*, int>> s;
s.push({root, 1});
while (!s.empty()) {
auto node = s.top();
s.pop();
if (!node.first->left && !node.first->right) {
depth = std::max(depth, node.second);
}
if (node.first->left) {
s.push({node.first->left, node.second + 1});
}
if (node.first->right) {
s.push({node.first->right, node.second + 1});
}
}
return depth;
}
bool isBalanced(TreeNode* root) {
if (!root || (!root->left && !root->right)) {
return true;
}
bool left = isBalanced(root->left);
bool right = isBalanced(root->right);
int leftDepth = dfs(root->left);
int rightDepth = dfs(root->right);
int val = std::abs(leftDepth - rightDepth);
return val > 1 ? false : true &&
left &&
right;
}
};