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 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).”
_______6______ / \ ___2__ ___8__ / \ / \ 0 _4 7 9 / \ 3 5
For example, the lowest common ancestor (LCA) of nodes 2
and 8
is 6
. Another example is LCA of nodes 2
and 4
is 2
, since a node can be a descendant of itself according to the LCA definition.
二叉搜索树的最低公共祖先节点。由于二叉搜索树的特性,即任何一个节点的左子树中的节点值都比该节点值小,任何一个节点的右子树中的节点值都比该节点值大。可以根据这个特性在二叉树中查找。
对于给定的两个节点指针p和q,先调整p和q,使p指向值较小的那个节点,q指向值较大的那个节点;
然后从根节点node开始遍历,如果q的值小于node的值,表示p和q都在node的左子树中,更新node为node->left;
如果p的值大于node的值,表示p和q都在node的右子树中,更新node为node->right;
否则表示找到的这个节点就是p和q的最低公共祖先节点。
runtime:40ms
/** * 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||p==NULL||q==NULL) return NULL; //使p保存值较小的节点,q保存值较大的节点 if(p->val > q->val) { /* TreeNode *tmp=p; p=q; q=tmp; */ //上面的代码可以直接写成下面的样子 swap(p,q); } TreeNode *result=root; while(true) { if(q->val < result->val) result=result->left; else if(p->val >result->val) result=result->right; else return result; } } };