樹中兩個節點的最低公共祖先
樹是二叉查詢樹的情況
題目來自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.
我們可以知道:
如果當前的節點的值比要查詢的兩個節點大的話,說明最低公共節點在當前節點的左子樹。
如果當前的節點的值比要查詢的兩個節點小的話,說明最低公共節點在當前節點的右子樹。
/**
* 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;
}
};
樹是普通二叉樹
題目來自於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;//如果當前節點為NULL說明走到了葉節點都沒有找到兩個節點中的其中一個
if (root == p || root == q)
return root;//如果當前節點為p,q之中的一個,那麼返回當前找到的節點中的一個
TreeNode *L = lowestCommonAncestor(root->left, p, q);//左子樹中是否能最先找到p,q中的一個節點
TreeNode *R = lowestCommonAncestor(root->right, p, q);
if (L && R)
return root; //如果當前節點左右節點都各找到一個,那麼返回當前節點
return L ? L : R; //只在左節點或者右節點找到一個,說明還有一個節點是在當前節點的下面
}
};