leetcode Lowest Common Ancestor of a Binary Tree

1、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 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 binary search tree: root = [6,2,8,0,4,7,9,null,null,3,5]

Example 1:

Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
Output: 6
Explanation: The LCA of nodes 2 and 8 is 6.

Example 2:

Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
Output: 2
Explanation: The LCA of nodes 2 and 4 is 2, 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 BST.

当树为二叉搜索树时，当所找两节点值都大于（都小于）当前节点值时，就在右子树（左子树）中递归查找；

否则两个节点值分别大于、小于当前节点值或有一个等于当前节点值，当前节点即为公共祖先。

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if (p->val > root->val && q->val > root->val)
            return lowestCommonAncestor(root->right, p, q);
        else if (p->val < root->val && q->val < root->val)
            return lowestCommonAncestor(root->left, p, q);
        else
            return root;
    }
};

2、binary tree

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

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

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.

当前节点为空或等于所找节点中的一个时，就返回当前节点；

否则分别在左子树和右子树中查找，若左右返回节点均非空时，由于树中无重复元素，所以所找两个节点分别在左子树和右子树中找到，当前节点就是公共祖先，返回当前节点。

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if (root == NULL || root == p || root == q)
            return root;
        auto l = lowestCommonAncestor(root->left, p, q);
        auto r = lowestCommonAncestor(root->right, p, q);
        if (l && r) return root;
        if (!l) return r;
        return l;
    }
};