Create Lowest_Common_Ancestor_of_a_Binary_Tree.js

Author: ignacio-chiazzo

LeetcodeProblems/Lowest_Common_Ancestor_of_a_Binary_Tree.js

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+/*
+Lowest Common Ancestor of a Binary Tree
+
+https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/
+
+Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
+
+According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”
+
+Given the following binary tree:  root = [3,5,1,6,2,0,8,null,null,7,4]
+
+        _______3______
+       /              \
+    ___5__          ___1__
+   /      \        /      \
+   6      _2       0       8
+         /  \
+         7   4
+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.
+Note:
+
+All of the nodes' values will be unique.
+p and q are different and both values will exist in the binary tree.
+*/
+
+var lowestCommonAncestor = function(root, p, q) {
+  if(root === null) {
+    return root;
+  }
+  if(p.val === root.val) {
+    return p;
+  }
+  if(q.val === root.val) {
+    return q;
+  }
+  if(lowestCommonAncestor(root.left, p, q));
+  const right = lowestCommonAncestor(root.right, p, q);
+  if(left !== null && right !== null) {
+    return root;
+  }
+  return left !== null ? left : right;
+};
+
+var lowestCommonAncestor1 = function(root, p, q) {
+  const pathToP = pathTo(root, p.val);
+  const pathToQ = pathTo(root, q.val); 
+  var elem = null;
+  for(var i = 0; i < Math.min(pathToP.length, pathToQ.length); i++) {
+    if(pathToP[i] !== pathToQ[i]) {
+      return pathToP[i - 1];
+    }
+  }
+    
+  return elem;
+};
+
+var pathTo = function(root, a) {
+    if(root === null) {
+        return null;
+    }
+    if(root.val === a) {
+        return [root.val];
+    }
+    const left = pathTo(root.left, a);
+    if (left !== null) {
+        return [root.val] + left;
+    }
+    const right = pathTo(root.right, a);
+     if(right !== null) {
+        return [root.val] + right;
+    }
+    return null;
+}
+
+