Amazon
  • Introduction
  • Phone Interview I
    • 53.Reverse Words in a String
    • 31.Partition Array
  • Phone Interview II
    • 167.Add Two Numbers
    • 88.Lowest Common Ancestor
  • Onsite I
    • 655.Big Integer Addition
    • 221.Add Two Numbers II
  • Onsite II
    • 158.Two Strings Are Anagrams
    • 171.Anagrams
    • 386.Longest Substring with At Most K Distinct Characters
  • Onsite III
    • 479.Second Max of Array
    • 589.Connecting Graph
  • Onsite IV
    • 532.Reverse Pairs
  • 2022
    • OA
      • work simulation
      • Greyness
      • NearestRetailer
      • Sum of Scores of Subarray
      • StrengthOfPassword
      • ProductOf1
      • Move 0/1 InArray
      • Max deviation among all substrings
      • AWS power consumption
      • searchWordResultWord
      • maxOperationOfString
      • MinHealthGame
      • EarliestMonth
      • Package ship
      • RatingConsectiveDecresing
      • LinkedListSum
      • MovingBoxes
      • ValidString
      • MaxValueAfterRemovingFromString
      • Subtree with Maximum Average
    • VO
      • 2022.3
    • BQ
      • doc
      • 2022.4
      • Freq Question
      • 11大类BQ和Follow-ups
      • Page 1
      • BQ 100
      • 35 behavioral questions asked in 95% of Amazon interviews with examples
      • Page 2
      • 反向BQ
    • LP
      • LP-1
      • LP-2
    • SD
      • Design Amazon Prime Video Recommendation System
      • Amazon Order system
    • OOD
      • Linux Find Command
      • Amazon Locker
    • AWS Identity call
    • Interviews
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  • Notice
  • 2.Code

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  1. Phone Interview II

88.Lowest Common Ancestor

1.Description(Medium)

Given the root and two nodes in a Binary Tree. Find the lowest common ancestor(LCA) of the two nodes.

The lowest common ancestor is the node with largest depth which is the ancestor of both nodes.

Notice

Assume two nodes are exist in tree.

Example

For the following binary tree:

  4
 / \
3   7
   / \
  5   6

LCA(3, 5) =4

LCA(5, 6) =7

LCA(6, 7) =7

2.Code

public TreeNode lowestCommonAncestor(TreeNode root, TreeNode A, TreeNode B) {
        if(root==null || root==A || root==B){//当前节点为空或者是AB其中一个就返回当前节点
            return root;
        }

        //Divided
        TreeNode left=lowestCommonAncestor(root.left,A,B);
        TreeNode right=lowestCommonAncestor(root.right,A,B);

        //Conquer
        if(left!=null && right!=null){  //AB位于左右子树两侧
            return root;
        }
        if(left!=null){            //此处右子树应该为空,因为找不到A,B任何一个,AB全在左子树上
            return left;
        }
        if(right!=null){
            return right;
        }

        return null;
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