236. 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.

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<TreeNode*> &path, vector<TreeNode *> &path1, vector<TreeNode*> &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<TreeNode*> 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;
    }
};