LeetCode OJ 236. Lowest Common Ancestor of a Binary Tree

最新推荐文章于 2022-02-12 22:47:33 发布

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最新推荐文章于 2022-02-12 22:47:33 发布

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原文链接：http://www.cnblogs.com/YuNanlong/p/6087606.html

本文探讨了在二叉树中寻找两个指定节点的最低公共祖先(LCA)的问题，并提供了一种递归解决方案。该算法首先检查根节点是否为空或是否为所求节点之一；然后递归地搜索左子树和右子树。如果左子树和右子树的搜索结果均不为空，则返回当前根节点作为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.

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解答

这题有可能会时间超限，就是因为第一遍做的时候没有将两个递归调用的结果用局部变量保存，导致了多次调用的情况出现……

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
struct TreeNode* lowestCommonAncestor(struct TreeNode* root, struct TreeNode* p, struct TreeNode* q) {
    struct TreeNode *left, *right;
    if(NULL == root){
            return NULL;
        }
        if(root == p||root == q){
            return root;
        }
        left = lowestCommonAncestor(root->left, p, q);
        right = lowestCommonAncestor(root->right, p, q);
        if(NULL == left){
            return right;
        }
        if(NULL == right){
            return left;
        }
        return root;
}