3715 · Lowest Common Ancestor VPRE
Algorithms
Medium
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Description
Given a binary tree with a root node root and an array nodes of objects of class TreeNode, you need to find the Lowest Common Ancestor (LCA, Lowest Common Ancestor) of all nodes in nodes and return it.
Where all the nodes in nodes exist in that binary tree and all the nodes in the binary tree have different values from each other.
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The number of nodes in the tree is in the range
[
1
,
1
0
4
]
[1,10
4
]
−
1
0
9
≤
�
�
�
�
�
�
�
�
.
�
�
�
≤
1
0
9
−10
9
≤TreeNode.val≤10
9
All TreeNode.val are unique
All nodes[i] are unique
All nodes[i] exist in the tree
Example
Example 1
Input
root = {3,5,1,6,2,0,8,#,#,7,4}
nodes = [4,7]
Output
2
Explanation
The lowest common ancestor of TreeNode(4) and TreeNode(7) is TreeNode(2)
Example 2
Input
root = {3,5,1,6,2,0,8,#,#,7,4}
nodes = [2]
Output
2
Explanation
The lowest common ancestor of a single node is the node itself
Example 3
Input
root = {3,5,1,6,2,0,8,#,#,7,4}
nodes = [7,2,6,4]
Output
5
Explanation
The lowest common ancestor of TreeNode(7)、TreeNode(2)、TreeNode(6) and TreeNode(4) is TreeNode(5)
Example 4
Input
root = {3,5,1,6,2,0,8,#,#,7,4}
nodes = [0,2,3,1,5,8,6,4,7]
Output
3
Explanation
nodes contains all the nodes in the tree, and the lowest common ancestor of all the nodes in the tree is the root node
Tags
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解法1：套用的labuladong的模板。
/*** Definition of TreeNode:* class TreeNode {* public:* int val;* TreeNode *left, *right;* TreeNode(int val) {* this->val = val;* this->left = this->right = NULL;* }* }*/class Solution {
public:/*** @param root: The root node of a binary tree.* @param nodes: An array of objects of class TreeNode.* @return: The lowest common ancestor of nodes.*/TreeNode* lowestCommonAncestor(TreeNode *root, vector<TreeNode*> &nodes) {if (!root) return NULL;unordered_set<TreeNode *> nodeSet;for (auto pNode : nodes) nodeSet.insert(pNode);TreeNode *res = helper(root, nodeSet);return res;}
private:TreeNode* helper(TreeNode *root, unordered_set<TreeNode *> &nodeSet) {if (!root) return NULL;if (nodeSet.find(root) != nodeSet.end()) return root;TreeNode *left = NULL, *right = NULL;left = helper(root->left, nodeSet);right = helper(root->right, nodeSet);if (left && right) return root;return left ? left : right;}
};
)
