这道题和314 Binary Tree Vertical Order Traversal类似,区别就是纵坐标的计算不同,这道题不存在有多个节点重合到一个纵坐标的情况。如果某个节点的位置是X,那么它的左子节点应该是2X,右子节点位置是2X+1
/**
* 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 widthOfBinaryTree(TreeNode* root) {
if(root==nullptr) return 0;
queue<pair<TreeNode*,long long int>> q;
q.push(make_pair(root,0));
long long int minCol=0;
long long int maxCol=0;
int maxWidth=1;
while(q.size()>0){
int size = q.size();
for(int i=0; i<size; i++){
pair<TreeNode*,int> cur = q.front();
q.pop();
TreeNode* node = cur.first;
long long int col = cur.second;
if(i==0){
minCol = col;
}
if(i==size-1){
maxCol = col;
maxWidth = max(maxWidth, maxCol-minCol+1);
}
if(node->left != nullptr){
q.push({node->left, col*2});
}
if(node->right != nullptr){
q.push({node->right, col*2+1});
}
}
}
return maxWidth;
}
};
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