235. Lowest Common Ancestor of a Binary Search Tree
2017-08-18 17:40
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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).”
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.
这里要充分利用二叉搜索时的性质,经过分析可以分以下几种情况:
2)节点一个在左,一个在右,或到最后其中一个根本就是根节点本身。
2)都在右子树中,需要在右子树中继续搜索公共祖先
根据上述情况,可以写出下面的递归函数进行求解:
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.
这里要充分利用二叉搜索时的性质,经过分析可以分以下几种情况:
终止条件
1)根节点直接为NULL;2)节点一个在左,一个在右,或到最后其中一个根本就是根节点本身。
递归过程
1)都在左子树中,需要在左子树中继续搜索公共祖先2)都在右子树中,需要在右子树中继续搜索公共祖先
根据上述情况,可以写出下面的递归函数进行求解:
class Solution { public: // 给定一棵二分搜索树,求两个节点的最近公共祖先 TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { // 递归终止条件 if (root == NULL || p == NULL || q == NULL) return NULL; // 找到公共祖先的情况 if (p->val <= root->val && q->val >= root->val || p->val >= root->val && q->val <= root->val) return root; // 左子树中递归 if (p->val < root->val && q->val < root->val) return lowestCommonAncestor(root->left, p, q); // 右子树中递归 if (p->val>root->val && q->val>root->val) return lowestCommonAncestor(root->right, p, q); } };
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