Lowest Common Ancestor of a Binary Tree(二叉樹的最近公共父親節點)

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

給定一個二叉樹和所要查詢的兩個節點，找到兩個節點的最近公共父親節點（LCA）。比如，節點5和1的LCA是3，節點5和4的LCA是5。

解題思路

本題與LeetCode 235的區別就是由二叉查詢樹變為了二叉樹，即資料從有序變為了無序，那麼就不能通過root的值和兩個節點的值的大小關係來劃分查詢區域。在題235的程式碼基礎做了些許變動，同樣使用遞迴搜尋的方法，當root非空時，對root->left和root->right分別進行搜尋。若搜尋結果均非空，說明兩個節點分別位於左右子樹之中，LCA則為root；若只有一個結果為空，則LCA是另一個非空的節點；若結果均空，則返回NULL。

c++程式碼如下：

/**
 * 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* left = lowestCommonAncestor(root->left,p,q);
        TreeNode* right = lowestCommonAncestor(root->right,p,q);
        if (left && right)
            return root;
        if (!left)
            return right;
        if (!right)
            return left;

    }
};