LeetCode 236. Lowest Common Ancestor of a Binary Tree--递归

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本文介绍了一种解决二叉树中寻找两个节点最低公共祖先(LCA)问题的方法。通过递归遍历二叉树，当根节点为空或为指定节点之一时返回该节点；若左右子树分别返回非空，则当前根节点即为最低公共祖先。

题目链接

236. Lowest Common Ancestor of a Binary Tree

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：TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q)

1.root是NULL返回NULL

2.root是p or q，返回root

3.root->left返回值不是NULL && root->right返回值不是NULL，说明root即为所求，返回root

4.root->left返回值不是NULL or root->right返回值不是NULL，说明不为NULL的返回值即为所求或为p，q，返回它

5.没找到p or q，返回NULL

代码：

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