[LeetCode] 1650. Lowest Common Ancestor of a Binary Tree III

Given two nodes of a binary tree p and q, return their lowest common ancestor (LCA).

Each node will have a reference to its parent node. The definition for Node is below:

class Node {
    public int val;
    public Node left;
    public Node right;
    public Node parent;
}

According to the definition of LCA on Wikipedia: "The lowest common ancestor of two nodes p and q in a tree T is the lowest node that has both p and q as descendants (where we allow a node to be a descendant of itself)."

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.

Example 3:

Input: root = [1,2], p = 1, q = 2
Output: 1

Constraints:

• The number of nodes in the tree is in the range [2, 105].
• -109 <= Node.val <= 109
• All Node.val are unique.
• p != q
• p and q exist in the tree.

二叉树的最近公共祖先 III。

给定一棵二叉树中的两个节点 p 和 q，返回它们的最近公共祖先节点（LCA）。

每个节点都包含其父节点的引用（指针）。Node 的定义如下：

class Node {
public int val;
public Node left;
public Node right;
public Node parent;
}
根据维基百科中对最近公共祖先节点的定义：“两个节点 p 和 q 在二叉树 T 中的最近公共祖先节点是后代节点中既包括 p 又包括 q 的最深节点（我们允许一个节点为自身的一个后代节点）”。一个节点 x 的后代节点是节点 x 到某一叶节点间的路径中的节点 y。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/lowest-common-ancestor-of-a-binary-tree-iii
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这道题跟前两个版本的区别是多了一个 parent 节点。这样我们就可以从当前节点反过来往回找父节点是谁。既然还是找两个节点的最小公共父节点，那么我们就从两个节点分别开始找他们各自的父节点。这里我首先去看一下两个节点的深度分别是多少，并把他们的深度先调整成一样。当深度一样的时候，方便两个节点同时往他们各自的父节点走，这样他们可以同时到达他们共同的父节点。

时间O(h)

空间O(1)

Java实现

/*
// Definition for a Node.
class Node {
    public int val;
    public Node left;
    public Node right;
    public Node parent;
};
*/

class Solution {
    public Node lowestCommonAncestor(Node p, Node q) {
        int pDepth = getDepth(p);
        int qDepth = getDepth(q);
        while (pDepth > qDepth) {
            pDepth--;
            p = p.parent;
        }
        while (pDepth < qDepth) {
            qDepth--;
            q = q.parent;
        }

        while (p != q) {
            p = p.parent;
            q = q.parent;
        }
        return p;
    }

    private int getDepth(Node node) {
        int depth = 0;
        while (node != null) {
            node = node.parent;
            depth++;
        }
        return depth;
    }
}

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1650. Lowest Common Ancestor of a Binary Tree III

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