Lowest Common Ancestor of a Binary Tree

Description

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).”

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.

Example 3:

Input: root = [1,2], p = 1, q = 2 Output: 1

Constraints:

• The number of nodes in the tree is in the range [2, 105].
• -109 <= Node.val <= 109
• All Node.val are unique.
• p != q
• p and q will exist in the tree.

Code

解題關鍵在於，對於一個 TreeNode 來說，如果自身就是其中之一，那自己就會是答案（因為是由上往下遞迴，在上面就可以 return 了）。如果自己不是且左右子樹分別都有找到，自己就會是答案（因為是由上往下遞迴，所以越下面的會越先得到結果然後 return ）。

/**
 * 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 || root == p || root == q) return root;
        TreeNode* Ancestor1 = lowestCommonAncestor(root->left, p, q);
        TreeNode* Ancestor2 = lowestCommonAncestor(root->right, p, q);
 
        if(Ancestor1 && Ancestor2) return root;
        return Ancestor1 ? Ancestor1 : Ancestor2;
    }
};

Source

• Lowest Common Ancestor of a Binary Tree - LeetCode