題目描述
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
Return 3, which is the length of the path [4,2,1,3] or [5,2,1,3].
Note: The length of path between two nodes is represented by the number of edges between them.
解析
按照常用方法計算一個節點的深度：max(depth of node.left, depth of node.right) + 1。在計算的同時，經過這個節點的路徑長度爲 1 + (depth of node.left) + (depth of node.right) 。搜索每個節點並記錄這些路徑經過的點數最大值，期望長度是結果 - 1。
class Solution {
public:
int ans;
int diameterOfBinaryTree(TreeNode* root) {
ans=1;
getHeight(root);
return ans-1;
}
int getHeight(TreeNode* root){
if(root==NULL) return 0;
int l=getHeight(root->left),r=getHeight(root->right);
ans=max(ans,l+r+1);
return max(l,r)+1;
}
};
類似題
124. Binary Tree Maximum Path Sum
Given a non-empty binary tree, find the maximum path sum.
For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The path must contain at least one node and does not need to go through the root.
Example 1:
Input: [1,2,3]
1
/ \
2 3
Output: 6
Example 2:
Input: [-10,9,20,null,null,15,7] -10 / \ 9 20 / \ 15 7 Output: 42
分析：這題的情況與上題有所不同，路徑不一定是越長值就越大，需要在找的過程中判斷路徑的值是大於0還是小於0，小於0就捨棄掉這一段。
class Solution {
public:
int ans=INT_MIN;
int maxPathSum(TreeNode* root) {
maxPathUptoDown(root);
return ans;
}
int maxPathUptoDown(TreeNode* root){
if(root==NULL) return 0;
int l=maxPathUptoDown(root->left);
int r=maxPathUptoDown(root->right);
int maxPath=root->val+max(0,l)+max(0,r);
ans=max(ans,maxPath);
return max(max(l,r),0)+root->val;
}
};