LeetCode 236 Lowest Common Ancestor of a Binary Tree解题报告

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.

题目的意思是:给两个二叉树(普通的二叉树)的节点，求这两个节点的最近公共祖先。

思路：因为二叉树是普通的，所以没有办法直接通过值去进行递归搜索，如果是二叉搜索树，则可以根据root->val>=p->val&&root->val<=q->val(q->val>p->val), 直接判断该节点是否是所求的LCA。
由于是普通的二叉树，只能直接进行判断：
1）如果该节点刚好是两个节点之一，则返回该节点
2) 如果不是，则进行递归搜索root->left和root->right

具体代码如下：

/**
 * 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:
    /*
        因为只是普通的二叉树，并不是二叉搜索树，只能使用搜索的方式直接进行搜索
        当返回的节点lca, rca分别不为空的时候，该点即为所求的最低祖先
    */
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(root==NULL||p==NULL||q==NULL)
        {
            return NULL;
        }
        if(root==p)
        {
            return root;
        }
        if(root==q)
        {
            return root;
        }
        TreeNode *lca=lowestCommonAncestor(root->left,p,q);
        TreeNode *rca=lowestCommonAncestor(root->right,p,q);
        /*
        如果两个都不为空，则表示该点就是所求的点
            a
          ／  \
         b     c
         其他节点返回必为空
        */
        if(lca!=NULL&&rca!=NULL)
        {
            return root;
        }
        /*
        如果lca为空，rca不为空，则返回rca
        如果rca为空，lca不为空，则返回lca
        如果两者都为空，返回null
        */
        return lca==NULL?rca:lca;
    }
};