Lowest Common Ancestor of a Binary Search Tree

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[/b]).”

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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.
来源： <https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/> 思路，先判断入口是否有非法输入。 1 如果某一个root==p || root == q，那么LCA肯定是root（因为是top down，LCA肯定在root所囊括的树上，而root又是p q其中一个节点了，那么另外一个节点肯定在root之下，那么root就是LCA），那么返回root 2 如果root<min(p, q)，那么LCA肯定在右子树上，那么递归 3 如果max(p, q)<root，那么LCA肯定在左子树上，那么递归 4 如果p<root<q，那么root肯定为LCA

/**

* 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==p||root==q)

return root;

if(p->val>q->val)

swap(p,q);

if(root->val > p->val && root->val < q->val)

return root;

TreeNode *node;

if(root->val > max(p->val,q->val))

node=lowestCommonAncestor(root->left,p,q);

if(root->val < min(p->val,q->val))

node=lowestCommonAncestor(root->right,p,q);

return node;

}

};

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来自为知笔记(Wiz)