LeetCode之Find Largest Value in Each Tree Row
2017-07-06 16:19
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本题的意思是给定一个二叉树,输出这棵树每层中最大的数。
我采用DFS的算法来解决此题。按照从左至右DFS的顺序遍历给定的二叉树,先将每层最左边的数当作当前每层的最大数依次加入到数组,然后在DFS的过程中,不断更新同层的最大数,当遍历完树时,更新完的数组即为所求。
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
private:
void dfs(TreeNode* root, vector<int>& max, int d) {
if(root == NULL){
return;
}
if((d == 0 && max.empty()) || d == max.size()) {
max.push_back(root -> val);
} else {
max[d] = max[d] > root -> val ? max[d] : root -> val;
}
dfs(root -> left, max, d + 1);
dfs(root -> right, max, d + 1);
}
public:
vector<int> largestValues(TreeNode* root) {
vector<int> max;
dfs(root, max, 0);
return max;
}
};
我采用DFS的算法来解决此题。按照从左至右DFS的顺序遍历给定的二叉树,先将每层最左边的数当作当前每层的最大数依次加入到数组,然后在DFS的过程中,不断更新同层的最大数,当遍历完树时,更新完的数组即为所求。
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
private:
void dfs(TreeNode* root, vector<int>& max, int d) {
if(root == NULL){
return;
}
if((d == 0 && max.empty()) || d == max.size()) {
max.push_back(root -> val);
} else {
max[d] = max[d] > root -> val ? max[d] : root -> val;
}
dfs(root -> left, max, d + 1);
dfs(root -> right, max, d + 1);
}
public:
vector<int> largestValues(TreeNode* root) {
vector<int> max;
dfs(root, max, 0);
return max;
}
};
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