树是二叉查找树的情况
题目来自LeetCode：https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/
Lowest Common Ancestor of a Binary Search Tree Total Accepted: 3402 Total Submissions: 8709 My Submissions Question Solution
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.
我们能够知道：
假设当前的节点的值比要查找的两个节点大的话，说明最低公共节点在当前节点的左子树。
假设当前的节点的值比要查找的两个节点小的话，说明最低公共节点在当前节点的右子树。
/**
* 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->val>p->val&&root->val>q->val)
return lowestCommonAncestor( root->left, p, q);
else if(root->val<p->val&&root->val<q->val)
return lowestCommonAncestor( root->right, p, q);
else
return root;
}
};树是普通二叉树
题目来自于LeetCode：https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/
Lowest Common Ancestor of a Binary Tree Total Accepted: 700 Total Submissions: 2270 My Submissions Question Solution
Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
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).”
_______3______
/
___5__ ___1__
/ /
6 _2 0 8
/
7 4
For example, the lowest common ancestor (LCA) of nodes 5 and 1 is 3. Another example is LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.
/**
* 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)
return NULL;//假设当前节点为NULL说明走到了叶节点都没有找到两个节点中的当中一个
if (root == p || root == q)
return root;//假设当前节点为p,q之一，那么返回当前找到的节点中的一个
TreeNode *L = lowestCommonAncestor(root->left, p, q);//左子树中能否最先找到p,q中的一个节点
TreeNode *R = lowestCommonAncestor(root->right, p, q);
if (L && R)
return root; //假设当前节点左右节点都各找到一个。那么返回当前节点
return L ? L : R; //仅仅在左节点或者右节点找到一个。说明另一个节点是在当前节点的以下
}
};