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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.

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Related Topics:
Tree

Similar Questions:

Solution 1.

// OJ: https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree-iii/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
    int getLength(Node *p) {
        int ans = 0;
        for (; p->parent; p = p->parent, ++ans);
        return ans;
    }
public:
    Node* lowestCommonAncestor(Node* p, Node * q) {
        int a = getLength(p), b = getLength(q);
        if (a < b) swap(a, b), swap(p, q);
        a -= b;
        while (a-- > 0) p = p->parent;
        while (p != q) p = p->parent, q = q->parent;
        return p;
    }
};