[Leetcode] 236. Lowest Common Ancestor of a Binary Tree 解题报告_lowest common ancestor iii magicbean-优快云博客

本文链接：https://blog.youkuaiyun.com/magicbean2/article/details/73848770

本文介绍了一种寻找二叉树中两个指定节点最低公共祖先的有效算法。通过递归搜索左右子树，若目标节点分布在不同子树则当前根节点即为所求。

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题目：

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.

思路：

相比于上一道题目，这道题目的难度确实增加了不少，当然解决方法也有不少。这里我给出一道在网上看到的非常巧妙的方法：我们分别在root的左右子树中搜索p和q。如果在左子树中搜索到了p或者q，就将其结点赋给left；如果在右子树中搜索到了p或者q，就将其结点赋给right。而如果没有搜索到，则返回NULL。这样如果left和right都不为NULL，则说明p和q必然位于不同的子树上，则root就是最低公共祖先；否则left和right中那个不为NULL的就是最低公共祖先（想一想是为什么呢？^_^答案是我们在递归的时候，返回的结点要么是p，要么是q，要么就是已经确定的p和q的最低公共祖先，请参考函数中的四个返回点进一步理解。）

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 || !p || !q) {
            return NULL;
        }
        if (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;
        }
        return left ? left : right;
    }
};