要想为true,要么左树或右树含p或q,要么当前节点是p或q return inl or inr or root == p or root == q def lowestCommonAncestor(self, root: TreeNode, p: TreeNode, q: TreeNode) -> TreeNode: if not root: return None # 注意python的实例变量必须定义在方法中 self.res = None self.dfs(...
public class Solution { public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { if (root == null) { return root; } if (root.val > Math.max(p.val, q.val)) { return lowestCommonAncestor(root.left, p, q); } else if (root.val < Math.min(p.val, q.val)...
https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/ 这里是BT不是BST,思路就是找到root到p和q的path然后找到最后的一个交点。要注意如何find Path。 如何用递归 http://bookshadow.com/weblog/2015/07/13/leetcode-lowest-common-ancestor-binary-tree/ 思路1 findPath。。。用post-orde...
LCA (Lowest Common Ancestor) 中文名称"最近公共祖先"。是指在树中,两个节点的最近的公共根。LeetCode里面的236题讲的就是这个。 实现 方法1:recursion publicTreeNodelca(TreeNode root,TreeNode p,TreeNode q){if(root==null||p==null||q==null){returnnull;}if(root==p||root==q){// 因为root是...
TreeNode236 right = lowestCommonAncestor(root.right,p,q); //如果左子节点和有子节点都成功返回了,说明匹配成功,答案就是当前节点 if(left == p && right == q || left == q && right == q) { return root; } } 如果左子树下没有节点能匹配到p或者q,就从右子树的根作为起点开始看向下递归,...
TreeNode l=lowestCommonAncestor(root.left,p,q); TreeNode r=lowestCommonAncestor(root.right,p,q);if(l!=null&&r!=null){returnroot; }if(l==null)returnr;returnl; } } 非递归: /** * Definition for a binary tree node. * public class TreeNode { ...
According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).” ...
12 TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { //这是一棵排序树啊! 13 if( (root->val - p->val ) * (root->val - q->val) <=0 ) //一大一小必会产生负的,或者root为其中1个,那么就是0
According to thedefinition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).” ...
精华题目列表: https://bit.ly/3fgcCiC 社群: turingplanet.org/turingplanet_community/ 视频纲要: 00:35 - 题目描述 01:54 - 解题思想 06:51 - 最优解代码解析相关系列:【刷题套路系列】https://bit.ly/3dgNzJL 【数据结构和算法入门】https://bit.ly/39cVLJ0 展开更多 知识 野生技能协会 LeetCode刷...