二叉树的最近公共祖先

  二叉树的最近公共祖先

 

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]

 

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.

 

 

解题的思路：

1. 如果输入的元素就是根 （root == q || root == p）则直接返回这个根

2. 如果输入的元素不是根 则root往下探一层看看是不是找到了其中的q或者p 如果找到了则说明这个root就是根

 

struct TreeNode* lowestCommonAncestor(struct TreeNode* root, struct TreeNode* p, struct TreeNode* q) {
    if(root == NULL)
        return NULL;
    if(root == p || root == q)
        return root;
    struct TreeNode* left = lowestCommonAncestor(root->left, p, q);
    struct TreeNode* right = lowestCommonAncestor(root->right, p, q);
    if(left != NULL && right != NULL)
        return root;
    else if(left!= NULL)
        return left;
    else if(right != NULL)
        return right;
    
    return NULL;    
    
}

请看程序， 

struct TreeNode* left = lowestCommonAncestor(root->left, p, q);
struct TreeNode* right = lowestCommonAncestor(root->right, p, q);
是在向下找有没有符合root == p 或者q的情况 如果找到了就是最小的那个根了。。
说实话 一开始真的没想到这种解法。脑子真的是不行了。。。哎。。