LeetCode543. Diameter of Binary Tree
Description
Given a binary tree, you need to compute the length of the diameter of the tree. The diameter of a binary tree is the length of the longest path between any two nodes in a tree. This path may or may not pass through the root.
Example:
Given a binary tree
1
/ \
2 3
/ \
4 5
Return3
, which is the length of the path[4,2,1,3]
or[5,2,1,3]
.
my program
思路:
int depth(TreeNode* root)
求树的高度,int depthDiff(TreeNode* root)
求root的左子树和右子树的高度和。递归求树的diameter
class Solution {
public:
int depth(TreeNode* root)
{
if (root == NULL) return 0;
return max(depth(root->left),depth(root->right))+1;
}
int depthDiff(TreeNode* root)
{
if(root == NULL) return 0;
return depth(root->left)+depth(root->right);
}
int diameterOfBinaryTree(TreeNode* root) {
if(root == NULL) return 0;
return max(depthDiff(root),max(diameterOfBinaryTree(root->left),diameterOfBinaryTree(root->right)));
}
};
Submission Details
106 / 106 test cases passed.
Status: Accepted
Runtime: 19 ms
Your runtime beats 18.82 % of cpp submissions.
other methods
C++ Solution with DFS
class Solution {
public:
int maxdiadepth = 0;
int dfs(TreeNode* root){
if(root == NULL) return 0;
int leftdepth = dfs(root->left);
int rightdepth = dfs(root->right);
if(leftdepth + rightdepth > maxdiadepth) maxdiadepth = leftdepth + rightdepth;
return max(leftdepth +1, rightdepth + 1);
}
int diameterOfBinaryTree(TreeNode* root) {
dfs(root);
return maxdiadepth;
}
};
C++_Recursive_with brief explanation
class Solution {
public:
int diameterOfBinaryTree(TreeNode* root) {
if(root == nullptr) return 0;
int res = depth(root->left) + depth(root->right);
return max(res, max(diameterOfBinaryTree(root->left), diameterOfBinaryTree(root->right)));
}
int depth(TreeNode* root){
if(root == nullptr) return 0;
return 1 + max(depth(root->left), depth(root->right));
}
};
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