# 236.Lowest Common Ancestor of a Binary Tree最近公共祖先  qlmx
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2020年2月15日00:19:25 评论 709阅读5分31秒

# 236. Lowest Common Ancestor of a Binary Tree(最近公共祖先)

## 1 问题描述

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 p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

Given the following binary tree: root = [3,5,1,6,2,0,8,null,null,7,4]

Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of nodes 5 and 1 is 3.


Example 2:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.


Note:

• All of the nodes' values will be unique.
• p and q are different and both values will exist in the binary tree.

## 2 C++

/**
* 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) {
vector<TreeNode*> path;
vector<TreeNode*> p_node;
vector<TreeNode*> q_node;
int finish = 0;
preorder(root, p, path, p_node, finish);
path.clear();
finish = 0;
preorder(root, q, path, q_node, finish);
int min_path_len = 0;
if(p_node.size() < q_node.size())
min_path_len = p_node.size();
else
min_path_len = q_node.size();

TreeNode* result = 0;
for(int i=0; i<min_path_len; i++) {
if(p_node[i] == q_node[i]) {
result = p_node[i];
}
}

return result;
}
private:
void preorder(TreeNode* node, TreeNode* search, vector<TreeNode*> &path, vector<TreeNode*> &result, int finish) {
if(!node || finish)
return;

path.push_back(node);

if(node == search) {
finish = true;
result = path;
}

preorder(node->left, search, path, result, finish);
preorder(node->right, search, path, result, finish);
path.pop_back();
}
};


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