236. 二叉树的最近公共祖先 - 力扣(LeetCode)
dfs统计根节点到p,q节点的路径,两条路径中最后一个相同节点就是公共祖先
/**
* 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 {
public:
bool dfs(TreeNode *root, TreeNode *p, vector<TreeNode*> &pp) {
if (root == nullptr) return false;
pp.push_back(root);
if (root == p) return true;
if (dfs(root->left, p, pp)) return true;
if (dfs(root->right, p, pp)) return true;
pp.pop_back();
return false;
}
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
vector<TreeNode*> p1, p2;
dfs(root, p, p1), dfs(root, q, p2);
int i, j;
for (i = 0, j = 0; i < p1.size() && j < p2.size(); ++ i, ++ j) {
if (p1[i] != p2[j]) break;
}
return p1[i - 1];
}
};
124. 二叉树中的最大路径和 - 力扣(LeetCode)
每次递归判断:以当前节点为起点的最长路径,需要先判断以左右子节点为根节点的最长路径(可能为0,表示路径中没有节点),然后加上当前节点值
同时维护“经过当前节点的最长路径”
/**
* 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 ans = INT_MIN;
int dfs(TreeNode* root) {
if (root == nullptr) return 0;
int lv = max(0, dfs(root->left));
int rv = max(0, dfs(root->right));
int cv = root->val + lv + rv;
ans = max(ans, cv);
return root->val + max(lv, rv);
}
int maxPathSum(TreeNode* root) {
dfs(root);
return ans;
}
};