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).”
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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 };