LeetCode 235:Lowest Common Ancestor of a Binary Search Tree
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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.
//解题思路: //依据BST的性质,左子树节点的值<根节点的值。右子树节点的值>根节点的值 //记当前节点为node,从根节点root出发 //若p与q分别位于node的两側,或当中之中的一个的值与node同样,则node为LCA //若p的值小于node的值,则LCA位于node的左子树 //否则,LCA位于node的右子树 class Solution { public: TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { while (root){ if( (p->val - root->val)*(q->val - root->val)<=0 ) return root; else if (p->val < root->val) return lowestCommonAncestor(root->left, p, q); else return lowestCommonAncestor(root->right, p, q); } } };
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