[LeetCode] Lowest Common Ancestor of a Binary Tree

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.

方法一:根据路径找LCA。

 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) {
13         if (root == NULL) return;
14         path.push_back(root);
15         if (root == p) path1 = path;
16         if (root == q) path2 = path;
17         //找到两个节点后就可以退出了
18         if (!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     }
23     
24     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;
28         int idx = 0;
29         while (idx < path1.size() && idx < path2.size()) {
30             if (path1[idx] != path2[idx]) break;
31             else res = path1[idx++];
32         }
33         return res;
34     }
35 };

 

方法二:节点a与节点b的公共祖先c一定满足:a与b分别出现在c的左右子树上(如果a或者b本身不是祖先的话)。

 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     TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
13         if (root == NULL) return NULL;
14         if (root == p || root == q) return root;
15         TreeNode *L = lowestCommonAncestor(root->left, p, q);
16         TreeNode *R = lowestCommonAncestor(root->right, p, q);
17         if (L && R) return root;
18         return L ? L : R;
19     }
20 };

 

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