[LeetCode] Lowest Common Ancestor of a Binary Tree 二叉树的最小共同父节点

 

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).”

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

For example, the lowest common ancestor (LCA) of nodes 5 and 1 is 3. Another example is LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

 

这道求二叉树的最小共同父节点的题是之前那道Lowest Common Ancestor of a Binary Search Tree 二叉搜索树的最小共同父节点的Follow Up。跟之前那题不同的地方是，这道题是普通是二叉树，不是二叉搜索树，所以就不能利用其特有的性质，所以我们只能在二叉搜索树中来搜索p和q，然后从路径中找到最后一个相同的节点即为父节点，我们可以用递归来实现，写法很简洁，代码如下：

 

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

 

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