二叉樹中任意兩個節點的最近公共祖先

public class Solution {  
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {  
        //發現目標節點則通過返回值標記該子樹發現了某個目標結點  
        if(root == null || root == p || root == q) return root;  
        //查看左子樹中是否有目標結點，沒有為null  
        TreeNode left = lowestCommonAncestor(root.left, p, q);  
        //查看右子樹是否有目標節點，沒有為null  
        TreeNode right = lowestCommonAncestor(root.right, p, q);  
        //都不為空，說明做右子樹都有目標結點，則公共祖先就是本身  
        if(left!=null&&right!=null) return root;  
        //如果發現了目標節點，則繼續向上標記為該目標節點  
        return left == null ? right : left;  
    }  
}  

思路：從根節點開始遍歷，如果node1和node2中的任一個和root匹配，那么root就是最低公共祖先。 如果都不匹配，則分別遞歸左、右子樹，如果有一個 節點出現在左子樹，並且另一個節點出現在右子樹，則root就是最低公共祖先.  如果兩個節點都出現在左子樹，則說明最低公共祖先在左子樹中，否則在右子樹。

感覺很奇妙。引申的問題

如果給定的不是二叉樹，而是二叉搜索樹呢？會比較簡單一點，如果是帶有指向父節點的指針的樹，可以轉化為兩個鏈表求交匯點的問題。

 235. Lowest Common Ancestor of a Binary Search Tree

Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.

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).”

        _______6______
       /              \
    ___2__          ___8__
   /      \        /      \
   0      _4       7       9
         /  \
         3   5

For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

 

/**
 * 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) {
        while((root -> val - p -> val)*(root -> val - q -> val) > 0 ){
            root = root -> val > p -> val ? root -> left : root -> right;
        }
        return root;
    }
};