[LeetCode] Lowest Common Ancestor of a Binary Tree
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Lowest Common Ancestor of a Binary Tree
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.
1
/*
*
2
* Definition for a binary tree node.
3
* struct TreeNode {
4
* int val;
5
* TreeNode *left;
6
* TreeNode *right;
7
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
8
* };
9
*/
10
class
Solution {
11
public
:
12
bool getPath(TreeNode *root, TreeNode *p, vector<TreeNode *> &path) {
13if (root == NULL) returnfalse;
14 path.push_back(root);
15if (root == p) returntrue;
16if (getPath(root->left, p, path) || getPath(root->right, p, path)) returntrue;
17 path.pop_back();
18returnfalse;
19 }
2021 TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
22 vector<TreeNode*> path1, path2;
23 getPath(root, p, path1);
24 getPath(root, q, path2);
25 TreeNode *res = root;
26int idx = 0;
27while (idx < path1.size() && idx < path2.size()) {
28if (path1[idx] != path2[idx]) break;
29else res = path1[idx++];
30 }
31return res;
32 }
33 };
跟上面一个意思。
1
/*
*
2
* Definition for a binary tree node.
3
* struct TreeNode {
4
* int val;
5
* TreeNode *left;
6
* TreeNode *right;
7
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
8
* };
9
*/
10
class
Solution {
11
public
:
12
void getPath(TreeNode *root, TreeNode *p, TreeNode *q, vector<TreeNode*> &path, vector<TreeNode *> &path1, vector<TreeNode*> &path2) {
13if (root == NULL) return;
14 path.push_back(root);
15if (root == p) path1 = path;
16if (root == q) path2 = path;
17//找到两个节点后就可以退出了18if (!path1.empty() && !path2.empty()) return;
19 getPath(root->left, p, q, path, path1, path2);
20 getPath(root->right, p, q, path, path1, path2);
21 path.pop_back();
22 }
2324 TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
25 vector<TreeNode*> path, path1, path2;
26 getPath(root, p, q, path, path1, path2);
27 TreeNode *res = root;
28int idx = 0;
29while (idx < path1.size() && idx < path2.size()) {
30if (path1[idx] != path2[idx]) break;
31else res = path1[idx++];
32 }
33return res;
34 }
35 };
原文：http://www.cnblogs.com/easonliu/p/4643873.html
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