LeetCode #236 - Lowest Common Ancestor of a Binary Tree

题目描述：

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]

        _______3______
       /              \
    ___5__          ___1__
   /      \        /      \
   6      _2       0       8
         /  \
         7   4

Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 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.

和Lowest Common Ancestor of a Binary Search Tree类似，由于不是BST，所以不能根据节点的值来判断，所以需要通过搜索查找来确定节点。

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(root==NULL||root==p||root==q) return root;
        bool left_p=search(root->left,p);
        bool right_p=search(root->right,p);
        bool left_q=search(root->left,q);
        bool right_q=search(root->right,q);
        if((left_p&&right_q)||(left_q&&right_p)) return root;
        else if(left_p&&left_q) return lowestCommonAncestor(root->left,p,q);
        else if(right_p&&right_q) return lowestCommonAncestor(root->right,p,q);
    }
    
    bool search(TreeNode* root, TreeNode* node)
    {
        if(root==NULL) return false;
        else if(root==node) return true;
        else return search(root->left,node)||search(root->right,node);
    }
};