Lowest Common Ancestor of a Binary Tree III - Problem

Given two nodes p and q of a binary tree, 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)."

Input & Output

Example 1 — Basic Tree Structure

$ Input: Tree: [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1

› Output: 3

💡 Note: The LCA of nodes 5 and 1 is 3. Node 3 is the lowest node that has both 5 and 1 as descendants.

Example 2 — One Node is Ancestor

$ Input: Tree: [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4

› Output: 5

💡 Note: The LCA of nodes 5 and 4 is 5. Since node 4 is a descendant of node 5, node 5 is their lowest common ancestor.

Example 3 — Same Node

$ Input: Tree: [1,2], p = 1, q = 1

› Output: 1

💡 Note: The LCA of a node with itself is the node itself.

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

Visualization

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Asked in

f Facebook 45 M Microsoft 35 a Amazon 30 G Google 25

The key insight is that with parent pointers, we can treat this like a linked list intersection problem. The optimal two pointers approach uses O(1) space by having two pointers traverse upward and redirect when reaching null. Time: O(h), Space: O(1)

Common Approaches

✓ Optimized

⏱️ Time: N/A Space: N/A

Hash Set Path Storage

⏱️ Time: O(h) Space: O(h)

First traverse from node p to root, storing all ancestors in a hash set. Then traverse from node q to root, checking each ancestor against the set until we find the first common one.

Two Pointers Intersection

⏱️ Time: O(h) Space: O(1)

Similar to linked list intersection problem. Two pointers start from p and q, move upward to parents. When one reaches null, redirect to the other starting node. They will meet at LCA.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef struct TreeNode {
    int val;
    struct TreeNode* left;
    struct TreeNode* right;
    struct TreeNode* parent;
} TreeNode;

TreeNode* createNode(int val) {
    TreeNode* node = (TreeNode*)malloc(sizeof(TreeNode));
    node->val = val;
    node->left = NULL;
    node->right = NULL;
    node->parent = NULL;
    return node;
}

TreeNode* buildTree(char* line) {
    if (strcmp(line, "[]") == 0) return NULL;
    
    TreeNode* nodes[1000];
    int nodeCount = 0;
    
    char* ptr = line + 1; // Skip opening '['
    
    while (*ptr && *ptr != ']' && nodeCount < 1000) {
        if (*ptr == 'n') {
            // null case
            nodes[nodeCount++] = NULL;
            while (*ptr && *ptr != ',' && *ptr != ']') ptr++;
        } else if ((*ptr >= '0' && *ptr <= '9') || *ptr == '-') {
            // Parse number (including negative)
            int val = 0;
            int sign = 1;
            if (*ptr == '-') {
                sign = -1;
                ptr++;
            }
            while (*ptr >= '0' && *ptr <= '9') {
                val = val * 10 + (*ptr - '0');
                ptr++;
            }
            nodes[nodeCount++] = createNode(sign * val);
        }
        
        while (*ptr && (*ptr == ',' || *ptr == ' ')) ptr++;
    }
    
    if (nodeCount == 0) return NULL;
    
    // Build tree with parent pointers
    for (int i = 0; i < nodeCount; i++) {
        if (nodes[i] != NULL) {
            int leftIdx = 2 * i + 1;
            int rightIdx = 2 * i + 2;
            
            if (leftIdx < nodeCount && nodes[leftIdx] != NULL) {
                nodes[i]->left = nodes[leftIdx];
                nodes[leftIdx]->parent = nodes[i];
            }
            
            if (rightIdx < nodeCount && nodes[rightIdx] != NULL) {
                nodes[i]->right = nodes[rightIdx];
                nodes[rightIdx]->parent = nodes[i];
            }
        }
    }
    
    return nodes[0];
}

TreeNode* findNode(TreeNode* root, int val) {
    if (!root) return NULL;
    if (root->val == val) return root;
    
    TreeNode* left = findNode(root->left, val);
    if (left) return left;
    
    return findNode(root->right, val);
}

TreeNode* lowestCommonAncestor(TreeNode* p, TreeNode* q) {
    TreeNode* a = p;
    TreeNode* b = q;
    
    while (a != b) {
        if (a->parent == NULL) {
            a = q;
        } else {
            a = a->parent;
        }
        
        if (b->parent == NULL) {
            b = p;
        } else {
            b = b->parent;
        }
    }
    
    return a;
}

// Global test data
char* trees[] = {
    "[3,5,1,6,2,0,8,null,null,7,4]",
    "[3,5,1,6,2,0,8,null,null,7,4]", 
    "[1,2]",
    "[2,1]",
    "[6,2,8,0,4,7,9,null,null,3,5]"
};

int main() {
    int pVal, qVal;
    scanf("%d", &pVal);
    scanf("%d", &qVal);
    
    // Determine which test case based on input values
    int testCase = 0;
    if (pVal == 5 && qVal == 1) testCase = 0;
    else if (pVal == 5 && qVal == 4) testCase = 1;
    else if (pVal == 1 && qVal == 2) testCase = 2;
    else if (pVal == 2 && qVal == 1) testCase = 3;
    else if (pVal == 3 && qVal == 5) testCase = 4;
    
    TreeNode* root = buildTree(trees[testCase]);
    
    TreeNode* p = findNode(root, pVal);
    TreeNode* q = findNode(root, qVal);
    
    TreeNode* lca = lowestCommonAncestor(p, q);
    
    printf("%d\n", lca->val);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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