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 p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

Given the following binary tree: root = [3,5,1,6,2,0,8,null,null,7,4]

二叉树

Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of nodes 5 and 1 is 3.
Example 2:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

Note:

All of the nodes' values will be unique.
p and q are different and both values will exist in the binary tree.

题意：给定一个二叉树, 找到该树中两个指定节点的最近公共祖先。

思路：
如果以root为根的子树中包含p和q，则返回它们的最近公共祖先
如果只包含p，则返回p
如果只包含q，则返回q
如果都不包含，则返回NULL
右子树情况：
1.右边也都不包含，则right=NULL，最终需要返回NULL
2.右边只包含p或q，则right=p或者q，最终需要返回p或q
3.右边同时包含p和q，则right是最近公共祖先，我们最终也需要返回最近公共祖先

/**
 * 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||root==p||root==q) 
            return root;
        TreeNode *left=lowestCommonAncestor(root->left,p,q);
        TreeNode *right=lowestCommonAncestor(root->right,p,q);
        if(left!=NULL&&right!=NULL) 
            return root;//如果p,q刚好在左右两个子树上
        if(left==NULL) 
            return right;//仅在右子树
        if(right==NULL) 
            return left;//仅在左子树
        return root;
    }
};