标签:org 题目 技术 公共祖先 style target int binary span
Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.
According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”
Given binary search tree: root = [6,2,8,0,4,7,9,null,null,3,5]
Example 1:
Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8 Output: 6 Explanation: The LCA of nodes2
and8
is6
.
Example 2:
Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4 Output: 2 Explanation: The LCA of nodes2
and4
is2
, since a node can be a descendant of itself according to the LCA definition.
题目大意:
给定一个二叉树,和二叉树中的两个节点,判断这两个节点的公共祖先。
理 解:
两个节点的两种位置,当前根节点的左右子树中或者两个节点是父子关系。
代 码 C++:
/** * 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->val == root->val || q->val == root->val) return root; TreeNode *left,*right; left = lowestCommonAncestor(root->left,p,q); right = lowestCommonAncestor(root->right,p,q); if(left!=NULL && right!=NULL) return root; else if(left==NULL) return right; else if(right==NULL) return left; else return NULL; } };
运行结果:
执行用时 :76 ms, 在所有C++提交中击败了35.68%的用户
letecode [235] - Lowest Common Ancestor of a Binary Search Tree
标签:org 题目 技术 公共祖先 style target int binary span
原文地址:https://www.cnblogs.com/lpomeloz/p/11023732.html