Lowest Common Ancestor of a Binary Tree Part I

Given a binary tree, find the lowest common ancestor of two given nodes in the tree.

       /              \
    ___5__          ___1__
   /      \        /      \
   6      _2       0       8
         /  \
         7   4

If you are not so sure about the definition of lowest common ancestor (LCA), please refer to my previous post: Lowest Common Ancestor of a Binary Search Tree (BST) or the definition of LCA here. Using the tree above as an example, the LCA of nodes 5 and 1 is 3. Please note that LCA for nodes 5 and 4 is 5.

Top-down or bottom-up? Consider both approaches and see which one is more efficient.

A Top-Down Approach (Worst case O(n2) ):
Let’s try the top-down approach where we traverse the nodes from the top to the bottom. First, if the current node is one of the two nodes, it must be the LCA of the two nodes. If not, we count the number of nodes that matches either p or q in the left subtree (which we call totalMatches). If totalMatches equals 1, then we know the right subtree will contain the other node. Therefore, the current node must be the LCA. If totalMatches equals 2, we know that both nodes are contained in the left subtree, so we traverse to its left child. Similar with the case where totalMatches equals 0 where we traverse to its right child.

What is the run time complexity of this top-down approach?

First, just for fun, we assume that the tree contains n nodes and is balanced (with its height equals to log(n) ). In this case, the run time complexity would be O(n). Most people would guess a higher ordered complexity than O(n) due to the function countMatchesPQ() traverses the same nodes over and over again. Notice that the tree is balanced, you cut off half of the nodes you need to traverse in each recursive call of LCA() function. The proof that the complexity is indeed O(n) is left as an exercise to the reader.

What if the tree is not necessarily balanced? Then in the worst case the complexity could go up to O(n2). Why? Could you come up with such case? (Hint: The tree might be a degenerate tree).

A Bottom-up Approach (Worst case O(n) ):
Using a bottom-up approach, we can improve over the top-down approach by avoiding traversing the same nodes over and over again.

We traverse from the bottom, and once we reach a node which matches one of the two nodes, we pass it up to its parent. The parent would then test its left and right subtree if each contain one of the two nodes. If yes, then the parent must be the LCA and we pass its parent up to the root. If not, we pass the lower node which contains either one of the two nodes (if the left or right subtree contains either p or q), or NULL (if both the left and right subtree does not contain either p or q) up.

Sounds complicated? Surprisingly the code appears to be much simpler than the top-down one.

The LCA problem had been studied extensively by many computer scientists. There exists efficient algorithms for finding LCA in constant time after initial processing of the tree in linear time. For the adventurous reader, please read this article for more details: Range Minimum Query and Lowest Common Ancestor in Topcoder.

Further Thoughts:
What if each node in the binary tree has a link to its parent? Could you devise a non-recursive approach without using extra space?

» Continue reading Lowest Common Ancestor of a Binary Tree Part II.

VN:F [1.9.22_1171]
Rating: 4.6/5 (109 votes cast)
Lowest Common Ancestor of a Binary Tree Part I, 4.6 out of 5 based on 109 ratings

53 thoughts on “Lowest Common Ancestor of a Binary Tree Part I

  1. Pingback: Lowest Common Ancestor of a Binary Tree Part II » i has 1337 code

  2. www

    “we assume that the tree contains n nodes and is balanced (with its height equals to log(n) ). In this case, the run time complexity would be O(n). ”

    i think this is similar as O(n) time in making a heap..the prove can be found in CLRS page135

    VA:F [1.9.22_1171]
      1. sandy

        find left count
        if left count == 2 => go left
        if left count ==1 => find right count (cant take it for granted that the other node is also present)
        if left count ==0 => find right count

        so worst case we traversed n nodes
        but we reduce the sub problem by half if both the nodes are present on left or right side

        so we get number of operations in each recursion as

        n + n/2 + n/4 + .. 1

        = n(1+1/2+1/4+… 1/2^lgn)
        as we know 1 +1/2 +1/4 +… inifinity = 2
        we get n*k where k>1 but <2
        so T(n) for top down approach = O(n)

        VA:F [1.9.22_1171]
  3. Sean

    One C++ question here:
    If you are having Node* root, Node* p here, how did you overload the “==” operator to directly use them like if(root == p)?

    VA:F [1.9.22_1171]
    1. 1337c0d3r Post author

      All pointers in C++ are just 32-bit unsigned integer values, so you don’t need to overload the “==” operator to do comparison.

      VN:F [1.9.22_1171]
      1. Sean

        I see, so the assumption here is that the nodes pointed by p and q should be exactly the same objects to be found in the binary tree.

        VA:F [1.9.22_1171]
  4. zorrowang

    I think there is a small bug in Top-Down Approach. According to the definition of LCA, it allows a node to be a descendant of itself. So when node p = q (p or q is not the root), the LCA should be itself. But your approach figures out the root as the LCA.

    VA:F [1.9.22_1171]
  5. learner

    I don’t think the bottom-up approach code is *correct*, in the sample tree, find LCA(root, 5, 9) returns 5, it should be null.

    VA:F [1.9.22_1171]
      1. Lee

        If both the nodes are already found in one of the subtrees of root, it does traversal again in vain. May be a static variable will be suffice to check, if subtree traversal is required or not

        VA:F [1.9.22_1171]
    1. ab123

      I think the input in this question is two nodes whose LCA we need to find. So, the case where one of the node is incorrect does not arise. Thus, i think the solution is correct only.

      VA:F [1.9.22_1171]
  6. Nagarjuna

    Will the bottom up approach work if there are duplicate elements in the tree, and we are looking for LCA of elements with one of them being duplicate?

    VA:F [1.9.22_1171]
  7. linda

    if p or q is not in the tree, this will still return the ancestor of the existing one

    we won’t touch both p and q because of this line

    “if (root == p || root == q) return root;”

    consider a tree like this

    / \
    3 4
    / \
    6 8

    let p = 4 q = 8, when we reach 4, we will return, and 8 will never be visited.

    VA:F [1.9.22_1171]
    1. Sanjay Pandey

      no this will work correctly .
      first 8 would we visited then as LCA(root->left, p, q); will go left then after traversing left subtree of 1 will visit right subtree and then following condn hlds
      if(L && R)
      return root

      VN:F [1.9.22_1171]
  8. Mridul

    The bottom up approach has one bug.
    For eg: let us assume the tree is
    inorder : 1 3 4 5 8 9 10 15
    preorder : 10 8 4 3 1 5 9 15

    try to find the lca(4,40) :- function will return 4 as the answer, which is incorrect as 40 is not at all in the tree.

    this error occurs whenever there will be one of p or q present in the tree and other is not, because in line 2, we check that if root’s value is either equal to p or q, then return and we didn’t check for the other number.

    I try to solve this problem…but can’t able to fix this…….so guys help me in fixing this issue…

    thanks 1337c03dr for ur nice solutioin.. 🙂

    VA:F [1.9.22_1171]
  9. Sanjay Pandey

    the bottom up approach won’t work if the binary tree has duplicate element.
    suppose there are two 2’s and LCA is to be calculated of (2,10) as —
    / \
    3 6
    / \
    2 2
    / \
    8 7
    it would return add of 3
    correct if i am wrong

    VN:F [1.9.22_1171]
  10. EddieZ

    @Mridul, to solve the problem of one of node is not in the tree

    int LCA(N* n, N* a, N* b, N*& head)
    if(!n || !a || !b) return 0;

    int l = LCA1(n->l, a, b, head);
    int r = LCA1(n->r, a, b, head);

    if(l==2 || r == 2)
    return 2;
    else if((l==1 && r == 1)
    ||( (l == 1 || r == 1 ) && (n == a || n == b)) )
    head = n;
    return 2;
    else if(n==a || n == b)
    return 1;
    return max(l,r);

    VA:F [1.9.22_1171]
  11. Hitman

    i think if the problem with the bottom up algo is just the case when one of the nodes doesn’t exist, then we can just do a pre check for the existance of both nodes in o(N), by calling the counting function of the top-down approach on the root, and checking if the return value is 2.

    VA:F [1.9.22_1171]
    1. zhong zhang

      For babycandy and Sanjay, maybe you guys misunderstand the problem.
      In the problem, we just consider the node itself(not the data element of the node), which is a pointer. In Sanjay’s example, it’s not duplicated. The two nodes with data element 2 are different nodes.

      VN:F [1.9.22_1171]
  12. Subramanian Ganapathy

    Top down worst case O(n)

    DFS and return the stack of nodes from bottom to root of all nodes in the path of each node{first adn second}.

    Then do a merge like procedure by popping off the excess elements of one stack and then successively popping until both the stacks have the same top.

    VA:F [1.9.22_1171]
  13. rohit

    My attempt though in C# this solves the extreme edge case of one of the nodes not present in tree while keeping botom up approach intact
    public Node LCA(Node Root,Noe P,Node Q,Ref bool Pbool,ref bool Qbool)
    if (Root==null||P==null||Q==null)

    L = LCA(Root.Left,P,Q, ref Pbool, ref Qbool);
    R = LCA(Root.Left,P,Q, ref Pbool, ref Qbool);
    Pbool = true; // identify if both the values are identified in the tree before spitting out LCA
    Return Root;

    if (Q ==Root)
    Qbool = true; // identify if both the values are identified in the tree before spitting out LCA
    Return Root;
    Return L==null?R:L;

    VA:F [1.9.22_1171]
  14. Gordon

    To 1337c0d3r:
    1. Assuming 2 nodes already in tree. This question has been asked by a lot of people not only here but also in the one for BST. You should put it clearly in your post that you make this assumption or better yet, do not make this assumption and provide code works for these cases.
    2. You are using the return value is null or not as an indicator if you found at least one value in one side of the tree. Simply change it to a structure that has a node and 2 boolean values, you can know if you found one node and which node or both nodes on one side of the tree. And you can avoid call the other side of the tree.

    VN:F [1.9.22_1171]
  15. Pingback: Lowest common ancestor of a Binary Search Tree « ..MindWrite..

  16. deb

    The following code takes care of all the cases.. Enjoy. 🙂 🙂

    VA:F [1.9.22_1171]
  17. Ming

    I have one simple approach which is based on post-order of binary tree traversing.

    struct tree {
    struct tree *l, *r;
    int key;

    static int n1 = 0, n2 = 0; //0: the given node not visited, 1: the given node visted
    static struct tree *a1, *a2; //give two nodes

    static void visit(struct tree *t)
    if (t == a1)
    n1 = 1;
    if (t == a2)
    n2 = 1;

    void lca(struct tree *root)
    int a1, a2;

    a1 = n1;
    a2 = n2;

    if (!root) return;


    if (!a1 && !a2 && n1 && n2) {
    printf(“lca of n1 & n2 is node %d\n”, root->key);

    VN:F [1.9.22_1171]
  18. Pingback: Leetcode笔记(14)Lowest Common Ancestor of a Binary Tree - JavaYu

  19. Sean

    The above solution of the top-down apporach has one bug in countMatchesPQ() when p and q are equal nodes. When p equals q, the LCA is p/q itself.
    However, the above top-down solution will return the parent of p/q. Thus, countMatchesPQ() should be modified as below.

    VA:F [1.9.22_1171]
  20. Pingback: 自底向上遍历树避免重复计算 | xiangcaohello

  21. XBYY

    The Bottom-up method will not work if one of the nodes is not present inside the tree, I think we can use a variable for flag to represent which node have been alread found.

    VN:F [1.9.22_1171]
  22. Pingback: LeetCode | techinterviewsolutions

  23. Pingback: LCA | moonstonelin

  24. Pingback: Lowest Common Ancestor – Lintcode Java | Welkin Lan

  25. すべての通り、腕時計は小さい不思議な彫刻家の創建アーティストウィラード維岡マイクロ。多くの細い点ともっと見るタイプ片、検査を私達の手で1型の文章はここで。この件が黒くて、

    すべての通り、腕時計は小さい不思議な彫刻家の創建アーティストウィラード維岡マイクロ。多くの細い点ともっと見るタイプ片、検査を私達の手で1型の文章はここで。この件が黒くて、彼を飛び出してそれからカードと彼の作品は明らかガラスを通して統合一面を例に、画像上の湖。同協会は明確には間違いない:それはAさんといいました。週知のトリックは、カードは、その魔法のように飛ぶ。 アクアノウティックコピース これは1つの執行はフランスの「激浪派の芸術家に敬礼」ロバート、誰が作った作品としての監督、彫刻家で、と事件のマスター。 http://www.newkakaku.com/gq1.htm

    VA:F [1.9.22_1171]

Leave a Reply

Your email address will not be published. Required fields are marked *

You may use the <code> tag to embed your code.