题目描述
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.
题目思路
- <剑指Offer>面试题68: 树中两个节点的最低公共祖先
代码 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:
bool findPath(TreeNode* root, TreeNode* t, vector& v){
bool flag = false;
// 在节点中加入节点
v.push_back(root);
// 如果找到节点，则flag=true，不再往下递归，开始返回
if(root == t){
flag = true;
}
// 左孩子不为NULL，且目前还没找到t，才能通过左子树递归
if(root->left!=NULL && flag==false){
flag = findPath(root->left, t, v);
}
// 右孩子不为NULL，且已经查完左子树，在左子树中没找到t，
// 才能沿着右子树递归查找，否则在左子树中找到t，则准备返回
if(root->right!=NULL && flag==false){
flag = findPath(root->right, t, v);
}
// 千万注意这一点，在找到节点t之后，不能再出数组了，
// 此时数组中保存的是根节点到t的路径(如果没有if则路径依次出栈了，程序出错)
if(flag == false){
v.pop_back();
}
return flag;
}
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
// 如果三个节点有一个NULL，则返回NULL
if(root == NULL || p == NULL || q == NULL){
return NULL;
}
vector v1;
vector v2;
// 第一步、找到根节点到两个节点路径
bool t1 = findPath(root, p, v1);
bool t2 = findPath(root, q, v2);
// t1 t2 都为 true 正面在树中找到 p q
if(t1==true && t2==true){
int i=0, j=0;
TreeNode* tt;
// 逐步向后比较，找到最后一个相同的节点，即为输入两个节点的最低公共祖先
while(i < v1.size() && j < v2.size() && v1[i]==v2[j]){
tt = v1[i];
i++;
j++;
}
return tt;
}
else{ // 在树中不存在 p q
return NULL;
}
}
};