236. Lowest Common Ancestor of a Binary Tree

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最新推荐文章于 2024-02-06 06:38:28 发布

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分类专栏： leetcode 文章标签： tree

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本文介绍了一种寻找二叉树中两个指定节点的最低公共祖先(LCA)的算法。通过递归方法，该算法计算当前节点及其左右子树中包含目标节点的数量，若数量大于等于2且未找到答案，则当前节点即为目标LCA。

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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.

找出二叉树中的两个节点p和q的共同的最低层的祖先，这里定义的祖先可以是自己本身。用递归的方法，计算当前结点及其左右子树中含p和q的个数，如果有两个，且答案还没出现，说明当前结点就是p和q的共同的最低层的祖先，标记为答案。其后的搜索就可以直接跳过。

代码：

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        lca(root, p, q);
        return res;
    }
    
private:
	TreeNode* res = NULL;
	int lca(TreeNode* root, TreeNode* p, TreeNode* q)
	{
		if(!root || res) return 0;
		int ret = (root == p ? 1 : 0) + (root == q ? 1 : 0);
		ret += lca(root->left, p, q);
		ret += lca(root->right, p, q);
		if(ret >= 2 && !res) res = root;
		return ret;
	}
};