Lowest Common Ancestor of a Binary 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.

抛开那些套路般的算法（下面会详细介绍）模板，这个题目也是可以做的。注意到它只是进行了一次询问，所以会比较好考虑。

注意到这样几点（假设所求节点为p和q）：

• 如果节点node为p和q的LCA的话，node的所有祖先都会是p和q的公共祖先
• 对于节点p和q来说，设他们的LCA为r，则要么p和q其中一个为r，要么他们分别处于r的左子树和右子树。

所以如果采用DFS（后序遍历）的搜索方式，一旦判断到某个节点符合上面第二条的性质，那么就能马上肯定这个节点就是所求LCA。复杂度不难分析，O(N)

对应代码如下：

/**
 * 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 *pp = NULL;
    TreeNode *qq = NULL;
    TreeNode *lca = NULL;

    void dfs(TreeNode *root, TreeNode *p, TreeNode *q)
    {
        if (lca != NULL) return;
        if (root->left != NULL)
        {
            dfs(root->left, p, q);
            if (pp == root->left) pp = root;
            if (qq == root->left) qq = root;
        }
        if (root->right != NULL)
        {
            dfs(root->right, p, q);
            if (pp == root->right) pp = root;
            if (qq == root->right) qq = root;
        }
        if (lca != NULL) return;
        if (root == p)    pp = root;
        if (root == q)    qq = root;
        if (pp != NULL && qq != NULL && pp == qq) lca = root;
    }

    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        dfs(root, p, q);
        return lca;
    }
};