235. Lowest Common Ancestor of a Binary Search Tree(LCA最低公共祖先)
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.
思路:因为是bst, 如果root,的值介于pq之间,root就是共同祖先。
root不是共同祖先,则递归地判断左右子树是不是公共祖先。
如果两个节点值都小于根节点,说明他们都在根节点的左子树上,我们往左子树上找
如果两个节点值都大于根节点,说明他们都在根节点的右子树上,我们往右子树上找
如果一个节点值大于根节点,一个节点值小于根节点,说明他们他们一个在根节点的左子树上一个在根节点的右子树上,那么根节点就是他们的最近公共祖先节点。
画个图看一下,比如要找0和5的最近公共祖先节点,
递归版:
class Solution { public: TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { if(root == NULL) return NULL; if(root == p || root == q) return root; if (root->val > p->val && root->val > q->val) return lowestCommonAncestor(root->left,p,q); if (root->val < p->val && root->val < q->val) return lowestCommonAncestor(root->right,p,q); return root; } };
循环版:
1 class Solution { 2 public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { 3 while((root.val-p.val)*(root.val-q.val)>0) 4 root = ((root.val-p.val)>0?root.left:root.right); 5 return root; 6 7 } 8 }