236. Lowest Common Ancestor of a Binary Tree

题目

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.

解析

思路：从根节点开始遍历，如果node1和node2中的任一个和root匹配，那么root就是最低公共祖先。 如果都不匹配，则分别递归左、右子树，如果有一个 节点出现在左子树，并且另一个节点出现在右子树，则root就是最低公共祖先.  如果两个节点都出现在左子树，则说明最低公共祖先在左子树中，否则在右子树。感觉很奇妙。引申的问题

• 如果给定的不是二叉树，而是二叉搜索树呢？会比较简单一点，如果是带有指向父节点的指针的树，可以转化为两个链表求交汇点的问题。

class Solution_236 {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {

        if (root==NULL||root==q||root==p)
        {
            return root;
        }

        TreeNode* left = lowestCommonAncestor(root->left, p, q);
        TreeNode* right = lowestCommonAncestor(root->right, p, q);

        if (left!=NULL&&right!=NULL)
        {
            return root;
        }

        return left == NULL ? right : left;
    }
};

•   236. Lowest Common Ancestor of a Binary Tree