Leetcode 236
思路:1.如果p或q就是根节点,那么LCA=p或q,返回根节点(递归出口)
2.分治
2.1 Divide:分别计算左字树和右子树的LCA
2.2 Conquer:如果左字树和右子树的计算结果均不为空,则根节点就是p,q的LCA;如果左不为空而右为空,则返回左子树的计算结果;
如果右不为空而左为空,则返回右子树的计算结果。如果左右子树均为空,则返回空指针
/** * 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: TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { if(root==NULL) return NULL; if(p==root||q==root) return root; TreeNode *left=lowestCommonAncestor(root->left,p,q); TreeNode *right=lowestCommonAncestor(root->right,p,q); if(left!=NULL&&right!=NULL) return root; if(left!=NULL) return left; if(right!=NULL) return right; return NULL; } };