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 p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

Given the following binary tree: root = [3,5,1,6,2,0,8,null,null,7,4]

Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of nodes 5 and 1 is 3.

Example 2:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

Note:

• All of the nodes’ values will be unique.
• p and q are different and both values will exist in the binary tree.

题目：二叉树的最近公共祖先。

思路：参见Discuss。遍历查找，如果遇到结点p或者q则返回。当在左子树遇到目标结点时，查找当前右子树是否存在另一个目标结点。如果分布在左右子树上，那么根结点为最近的公共祖先；如果右子树始终找不到q，那么说明p是q的祖先结点。

工程代码下载

/**
 * 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 == nullptr || root == p || root == q)
            return root;
        TreeNode* left = lowestCommonAncestor(root->left, p, q);
        TreeNode* right = lowestCommonAncestor(root->right, p, q);
        if(left != nullptr && right != nullptr)
            return root;
        return left == nullptr ? right : left;
    }
};

标签：Lowest,LeetCode236,right,TreeNode,二叉树,nodes,null,root,left
来源： https://blog.csdn.net/grllery/article/details/96751890