So basically we have two values B & C. And we need to find the LCA of the nodes represented by these two values. There is no guarantee that B & C exist in the tree. In case any value is missing, we should return -1 from the method.

```
_______10______
/ \
___6__ ___11__
/ \ / \
17 _2_ 7 8
/ \
12 4
```

For the solution, we will be traversing the tree twice, one for B & one for C. We will store paths to B & C separately in ArrayList variables. In case B or C is missing, that path will be empty. Once we have calculated the path, we will compare the paths & find the first element where there is a diversion or the value is equal to B or C. That will give us the LCA of the binary tree for these two values. Time complexity of this solution is O(n).

Here is a working Java code implementation of this problem.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 | import java.util.*; import java.lang.*; import java.io.*; //Definition for binary tree class TreeNode { int val; TreeNode left; TreeNode right; TreeNode(int x) { val = x; left=null; right=null; } } class Solution { public int lca(TreeNode A, int B, int C) { List<Integer> path1 = new ArrayList<Integer>(); List<Integer> path2 = new ArrayList<Integer>(); TreeNode temp = A; findPathToValue(temp, path1, B); //resetting starting index temp = A; findPathToValue(temp, path2, C); if(path1.size() == 0 || path2.size() == 0){ return -1; } int smallerSize = Math.min(path1.size(), path2.size()); int i=0; for(;i<smallerSize; i++){ if(!path1.get(i).equals(path2.get(i))){ break; } } return path1.get(i - 1); } private boolean findPathToValue(TreeNode temp, List<Integer> path, int value){ path.add(temp.val); if(temp.val == value){ return true; } if(temp.left != null && findPathToValue(temp.left, path, value)){ return true; } if(temp.right != null && findPathToValue(temp.right, path, value)){ return true; } path.remove(path.size() - 1); return false; } } |

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