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.

经典问题！

方法一：找到两个节点的路径，然后根据路径找LCA。

/** * 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:    void getPath(TreeNode *root, TreeNode *p, TreeNode *q, vector
     
       &path, vector
      
        &path1, vector
       
         &path2) {        if (root == NULL) return;        path.push_back(root);        if (root == p) path1 = path;        if (root == q) path2 = path;        //找到两个节点后就可以退出了        if (!path1.empty() && !path2.empty()) return;        getPath(root->left, p, q, path, path1, path2);        getPath(root->right, p, q, path, path1, path2);        path.pop_back();    }    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {        vector
        
          path, path1, path2;        getPath(root, p, q, path, path1, path2);        TreeNode *res = root;        int idx = 0;        while (idx < path1.size() && idx < path2.size()) {            if (path1[idx] != path2[idx]) break;            else res = path1[idx++];        }        return res;    }};
        
       
      
     

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）。

/** * 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:    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {        if (root == NULL) return NULL;        if (root == p || root == q) return root;        TreeNode *L = lowestCommonAncestor(root->left, p, q);        TreeNode *R = lowestCommonAncestor(root->right, p, q);        if (L && R) return root;        return L ? L : R;    }};