Create 1740-find-distance-in-a-binary-tree.js

Author: everthis

1740-find-distance-in-a-binary-tree.js

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+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val, left, right) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.left = (left===undefined ? null : left)
+ *     this.right = (right===undefined ? null : right)
+ * }
+ */
+/**
+ * @param {TreeNode} root
+ * @param {number} p
+ * @param {number} q
+ * @return {number}
+ */
+const findDistance = function (root, p, q) {
+  if (p == q) return 0
+  let result = -1
+  dfs(root, p, q)
+  return result
+
+  /**
+		The return value means the distance from root node to EITHER p OR q. If
+		neither p nor q are reachable from the root, return -1.
+		
+		It is either p or q but not both, because if the root node can reach both 
+		p and q, it is a common ancestor of p and q and the answer should already 
+		be available.
+	**/
+  function dfs(root, p, q) {
+    if (root == null) return -1
+
+    let left = dfs(root.left, p, q)
+    let right = dfs(root.right, p, q)
+
+    if (root.val == p || root.val == q) {
+      // root is p or q, but none of p or q is a descendent of root.
+      // The distance from root to one of p and q is 0 in this case.
+      if (left < 0 && right < 0) {
+        return 0
+      }
+
+      // root is p or q, and root is also the LCA of p and q.
+      result = 1 + (left >= 0 ? left : right)
+      return -1
+    }
+
+    // root is neither p nor q, but it is the LCA of p and q.
+    if (left >= 0 && right >= 0) {
+      result = left + right + 2
+      return -1
+    }
+
+    if (left >= 0) {
+      return left + 1
+    }
+
+    if (right >= 0) {
+      return right + 1
+    }
+
+    return -1
+  }
+}