236. Lowest Common Ancestor of a Binary Tree

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本文介绍了一种寻找二叉树中两给定节点的最低公共祖先(LCA)的有效算法。通过递归遍历左右子树并计数目标节点出现次数来定位LCA。示例说明了如何实现这一算法。

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题目描述

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).”
(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.

解题思路

如果找到该节点，则返回1，没有找到返回0，找到左边+右边=2的第一个点。因为其他公共子节点可能是左子树或者右子树为2，另一个子树为0。所以必须排除这种情况。我的出错地方在于应该判断两个节点相同，而不是判断两个节点的值相同。

代码

    TreeNode* lowestCommonAncestor(TreeNode* root,TreeNode* p, TreeNode* q) 
    {
        if(!root) return nullptr;
        TreeNode* ans = nullptr;
        int count = 0;
        CommonAncestor(root, p, q, ans, count);
        return ans;
    }
    void CommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q, TreeNode*& ans, int& count)
    {
        if(!root) return;
        int lCount = 0, rCount = 0, curCount = 0;
        if(root == p || root == q) curCount = 1;
        CommonAncestor(root->left, p, q, ans, lCount);
        CommonAncestor(root->right, p, q, ans, rCount);
        count = curCount + lCount + rCount;
        if(lCount < 2 && rCount < 2 && count == 2)
        {
            ans = root;
            return;
        }   
    }