Fix findDOMNode algorithm

src/renderers/shared/fiber/ReactFiberTreeReflection.js

@@ -101,17 +101,16 @@ function findCurrentFiberUsingSlowPath(fiber : Fiber) : Fiber | null {
   let b = alternate;
   while (true) {
     let parentA = a.return;
-    let parentB = b.return;
+    let parentB = parentA ? parentA.alternate : null;
     if (!parentA || !parentB) {
       // We're at the root.
       break;
     }
+
+    // If both copies of the parent fiber point to the same child, we can
+    // assume that the child is current. This happens when we bailout on low
+    // priority: the bailed out fiber's child reuses the current child.
     if (parentA.child === parentB.child) {
-      // If both parents are the same, then that is the current parent. If
-      // they're different but point to the same child, then it doesn't matter.
-      // Regardless, whatever child they point to is the current child.
-      // So we can now determine which child is current by scanning the child
-      // list for either A or B.
       let child = parentA.child;
       while (child) {
         if (child === a) {
@@ -133,8 +132,56 @@ function findCurrentFiberUsingSlowPath(fiber : Fiber) : Fiber | null {
         'Unable to find node on an unmounted component.'
       );
     }
-    a = parentA;
-    b = parentB;
+
+    if (a.return !== b.return) {
+      // The return pointer of A and the return pointer of B point to different
+      // fibers. We assume that return pointers never criss-cross, so A must
+      // belong to the child set of A.return, and B must belong to the child
+      // set of B.return.
+      a = parentA;
+      b = parentB;
+    } else {
+      // The return pointers pointer to the same fiber. We'll have to use the
+      // default, slow path: scan the child sets of each parent alternate to see
+      // which child belongs to which set.
+      //
+      // Search parent A's child set
+      let didFindChild = false;
+      let child = parentA.child;
+      while (child) {
+        if (child === a) {
+          didFindChild = true;
+          a = parentA;
+          b = parentB;
+          break;
+        }
+        if (child === b) {
+          didFindChild = true;
+          b = parentA;
+          a = parentB;
+          break;
+        }
+        child = child.sibling;
+      }
+      if (!didFindChild) {
+        // Search parent B's child set
+        child = parentB.child;
+        while (child) {
+          if (child === a) {
+            a = parentB;
+            b = parentA;
+            break;
+          }
+          if (child === b) {
+            b = parentB;
+            a = parentA;
+            break;
+          }
+          child = child.sibling;
+        }
+      }
+    }
+
     invariant(
       a.alternate === b,
       'Return fibers should always be each others\' alternates.'