236. Lowest Common Ancestor of a Binary Tree

栏目: 数据库 · 发布时间: 7年前

内容简介: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 des

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]

236. Lowest Common Ancestor of a Binary Tree

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.

难度:medium

题目:给定一二叉树,找出给定的两个结点的最低层的共同祖先结点。

根据LCA维基百科定义,最低公共祖先定义为两个节点p和q之间的最低公共祖先,它是T中同时具有p和q作为子代的最低节点(我们允许一个节点作为自身的子代)

思路:递归,后续遍历。

Runtime: 5 ms, faster than 100.00% of Java online submissions for Lowest Common Ancestor of a Binary Tree.

Memory Usage: 33.9 MB, less than 69.83% of Java online submissions for Lowest Common Ancestor of a Binary Tree.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null || root == p || root == q) {
            return root;
        }
        
        TreeNode left = lowestCommonAncestor(root.left, p, q);
        TreeNode right = lowestCommonAncestor(root.right, p, q);
        
        return left == null ? right : right == null ? left : root;
    }
}

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