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 };