Leetcode 236. Lowest Common Ancestor of a Binary Tree

最新推荐文章于 2022-10-10 12:58:14 发布

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最新推荐文章于 2022-10-10 12:58:14 发布

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分类专栏： Leetcode C++ 文章标签： Leetcode cplusplus

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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，在左右子树中分别查找目标节点，最终确定最低公共祖先的位置。

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原题：

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.

解决方法：

递归的办法。

- 如果root是其中之一，返回root.

- 分别从左右子树中找，如果左右子树各找其一，则返回root;如果在左子树没有找到，则返回在右子树找到的结果；如果在右子树没有找到，则返回在左子树找到的结果。

代码：

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;
        else if (!left)
            return right;
        else
            return left;
    }